I NEED HELP WITH THIS MATH PROBLEM!!!

\((8.2 \times 10^(^-^1^0^))\)\((5.3\times 10^(^-^9^))\)The sum of \((5.3 \times 10^(^-^9^))\) and \((8.2 \times 10^(^-^1^0^))\) in scientific notation is 1.35 x 10^−8.

To find the sum of \((5.3 \times 10^(^-^9^))\) and \((8.2 x 10^(^-^1^0^))\), we can add the coefficients and keep the same base, which is 10. Adding 5.3 and 8.2 gives us 13.5. Since both numbers are expressed in scientific notation, we need to adjust the decimal point to have one digit to the left of it.

The exponent in scientific notation represents the number of decimal places we need to move the decimal point to the left (for negative exponents) or to the right (for positive exponents). In this case, the exponents are -9 and -10.

Since -9 is larger than -10, we need to adjust the decimal point by 1 place to the left. Therefore, the sum of \((5.3 x 10^(^-^9^))\) and \((8.2 \times 10^(^-^1^0^))\) in scientific notation is \(1.35 \times 10^-^8^\).

Note: Scientific notation is a concise way of representing very large or very small numbers by using powers of 10. It consists of a coefficient (a decimal number between 1 and 10) multiplied by 10 raised to an exponent.

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Answer:

1. " second picture"

2. The graph of h(x) is shown below.

Step-by-step explanation:

The given function is

h(x)=7\sin xh(x)=7sinx

The general form of sine function is

f(x)=a\sin(bx+c)+df(x)=asin(bx+c)+d

Where, a is amplitude, b is period, c is phase shift and d is vertical shift.

So, the amplitude of the given function is 7, period is 1, phase shift is 0 and vertical shift is 0.

It means the minimum value of function is -7 and maximum value is 7.

Put x=0 in the given function.

h(x)=7\sin (0)=7(0)=0h(x)=7sin(0)=7(0)=0

Put x=-\frac{\pi}{2}x=− 2ππ in the given function.

h(x)=7\sin (-\frac{\pi}{2})=-7(1)=-7h(x)=7sin(−2π )=−7(1)=−7

Put x=\frac{\pi}{2}x= 2π in the given function.

h(x)=7\sin (\frac{\pi}{2})=7(1)=7h(x)=7sin( 2π )=7(1)=7

Therefore the points on the function are (0,0), (-\frac{\pi}{2},-7),(\frac{\pi}{2},7)(− 2π ,−7),( 2π,7) .

The graph of function is shown below.

"Hope this helps"

Answer:

Step-by-step explanation:

maximum is at 7, minimum is at -7, midline is at pie

If (x + 8) is a factor of f(x), then the true statement is f(-8) = 0 .

In the question ,

it is given that , x \(+\) 8 is factor of f(x) ,

we have to select the correct option about f(x) from the given options .

When solving the polynomials, a factor of the polynomial gives the solution of  the polynomial.

that means , x + 8 = 0

So , x = -8 ,

So , it means that when x = -8 , then the value of the function f(-8) will be zero ,

which means that x = -8 will be a root of the given function f(x) .

Therefore , the true statement is f(-8) = 0 .

The given question is incomplete , the complete question is

If (x + 8) is a factor of f(x), which of the following must be true ?

(a) Both x = –8 and x = 8 are roots of f(x).

(b) Neither x = –8 nor x = 8 is a root of f(x).

(c) f(–8) = 0

(d) f(8) = 0

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The experimental probability of tossing a 3 is 22% ( option D).

Experimental probability, which is also known as Empirical probability, is based on actual experiments and adequate recordings of the occurring of events. An experiment can be repeated a fixed number of times and each repetition is known as a trial. The formula for the experimental probability is defined by;

Probability of an Event P(E) = Number of times an event occurs / Total number of events

From the table it is visible that the occurrence of tossing 3 is 11 times.

So by the definition of experimental probability the possibility is 11/50

As the table given is the result for tossing a number cube 50 times.

In 50 times the possibility is 11

In 1 time the possibility is 11/50

In 100 times the possibility is (11×100)/50

                                              = 22

Hence, the experimental probability of tossing a 3 is 22%.

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Using the shell method, the volume of the solid generated when R is revolved about the y-axis is V= 324π cubic units.

The shell method refers to a method of finding volumes of the solid by decomposing a solid of revolution into cylindrical shells. It is one of the methods to find the volume of the solid. This solid is generated when region R is rotated about an axis or the y-axis. In case it is y-axis then the following formula is used:

V= ∫_a^b▒〖2πx [f(x)-g(x)]dx〗

Where y = f(x), y = g(x), and the interval is considered as [a,b].

The given curves are: y = 11 − x, y = 0, x = 3, x = 9

V= ∫_3^9▒〖2πx [11-x-0]dx〗

V= 2π∫_3^9▒〖(11x-x^2)dx〗

V= 2π[11/2 x^2-  1/3 x^3]

V= 2π[11/2 9^2-  1/3 9^3-  11/2 3^2+  1/3 3^3]

V= 2π[891/2-  729/3-  99/2+  27/3]

V= 2π[729/2-  702/3]

V= 2π[396-234]

V= 2π[162]

V= 324π cubic units

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Answer:

Step-by-step explanation:

In the Intermediate Phase of mathematics education, one topic that demonstrates a hierarchical and logical progression in patterns, functions, and algebra is the concept of "Linear Equations."

The topic of Linear Equations in the Intermediate Phase builds upon the foundation laid in earlier grades and serves as a stepping stone towards more advanced algebraic concepts. Here is an overview of the topic sequence and content area spread for Linear Equations:

Introduction to Variables and Expressions:

Students are introduced to the concept of variables and expressions, learning to represent unknown quantities using letters or symbols. They understand the difference between constants and variables and learn to evaluate expressions.

Solving One-Step Equations:

Students learn how to solve simple one-step equations involving addition, subtraction, multiplication, and division. They develop the skills to isolate the variable and find its value.

Solving Two-Step Equations:

Building upon the previous knowledge, students progress to solving two-step equations. They learn to perform multiple operations to isolate the variable and find its value.

Writing and Graphing Linear Equations:

Students explore the relationship between variables and learn to write linear equations in slope-intercept form (y = mx + b). They understand the meaning of slope and y-intercept and how they relate to the graph of a line.

Systems of Linear Equations:

Students are introduced to the concept of systems of linear equations, where multiple equations are solved simultaneously. They learn various methods such as substitution, elimination, and graphing to find the solution to the system.

Word Problems and Applications:

Students apply their understanding of linear equations to solve real-life word problems and situations. They learn to translate verbal descriptions into algebraic equations and solve them to find the unknown quantities.

The content area spread for Linear Equations includes concepts such as variables, expressions, equations, operations, graphing, slope, y-intercept, systems, and real-world applications. The progression from simple one-step equations to more complex systems of equations reflects a logical sequence that builds upon prior knowledge and skills.

By following this hierarchical progression, students develop a solid foundation in algebraic thinking and problem-solving skills. They learn to apply mathematical concepts and procedures in a systematic and logical manner, paving the way for further exploration of patterns, functions, and advanced algebraic topics in later phases of mathematics education.

The graph with the same end behavior is the one in option C.

The given function has a numerator with a negative leading coefficient and an odd degree, while the denominator is a quadratic polynomial.

It will mean that as x tends to negative infinity, the function tends to infinity, and as x tends to positive infinity, the function tends to negative infinty, that is the end behavior of the given function.

The graph with that end behavior is the same one than in option C.

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The recursive sequence that represents the sequence defined by the following explicit formula is given as follows:

A geometric sequence is a sequence of numbers where each term is obtained by multiplying the previous term by a fixed number called the common ratio q.

The explicit formula of the sequence is given as follows:

\(a_n = a_1q^{n-1}\)

In which \(a_1\) is the first term.

The first term and the common ratio for this problem is given as follows:

\(a_1 = -32, q = -\frac{1}{4}\)

The common ratio means that each term is the previous term multiplied by -1/4, hence the recursive formula is given as follows:

\(a_{n + 1} = -\frac{1}{4}a_n\)

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Answer:

2222222222222222222222222222222

Step-by-step explanation:

Answer:

A. AAS

Step-by-step explanation:

In both triangles, the line mn and jh are congruent, the angles k and g are congruent, and h and m are congruent.

The fraction that represents the rate between PKR and Paisas is given as follows:

1/50.

A fraction is a numerical representation of the division of the two terms x and y, as follows:

Fraction = x/y.

As the rate is PKR 1 = 50 paisas, we have that the fraction that represents the rate between PKR and Paisas is given as follows:

1/50.

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Answer:

4

Step-by-step explanation:

The area is 16, and you know that it's 2 sides u need to multiply

The perimeter is 16, and you know that you need to add up all the sides

So....

A = b * h

A = 4 * 4

A = 16    CORRECT

And....

P = S + S + S + S                                                         ( s = sides)

P = 4 + 4 + 4 + 4

P = 16 CORRECT

Hope this helped!

Have a supercalifragilisticexpialidocious day!

Answer:

-6y -8 = -18

18-8=6y

10=6y

y=10/6=5/3

Answer:

-6y - 8 = -18

     + 8    +8

-------------------

-6y = -10

-----    -----

-6       -6

10 divided by 6 is 5/3

so y is 5/3

m and n are the two numbers that form the fraction which is the probability of a randomly chosen positive divisor of \(10^99\) being an integral multiple of \(10^88\).

Let the probability of a randomly chosen positive divisor of \(10^99\)being an integral multiple of 10^88 be represented as m/n.

Since \(10^99\)is a multiple of \(10^88\), its positive divisors will also be multiples of \(10^88\).

Therefore, the probability that a randomly chosen divisor of \(10^99\)is an integral multiple of \(10^88\) will be 1, i.e., m/n = 1.

Therefore, m+n = 1+1 = 2

Hence, the value of (m+n) is 2.

m and n are two numbers that form a fraction in lowest terms, which is the probability that a randomly chosen positive divisor of \(10^99\) is an integral multiple of \(10^88\). To find the value of (m+n), we need to first determine the value of m and n. To do this, we must first calculate the total number of positive divisors of \(10^99\)that are integral multiples of \(10^88\). We can then divide this number by the total number of positive divisors of \(10^99\) to get the fraction. The numerator of this fraction is m and the denominator is n. The sum of m and n is the answer to the question. This is because the fraction gives us the probability that a randomly chosen positive divisor of \(10^99\) is an integral multiple of \(10^88\), and the sum of m and n is the total number of positive divisors of\(10^99\), which is the denominator of the fraction. Therefore, the value of (m+n) is the sum of the numerator and denominator of the fraction.

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According to this typical corporate bond listing of New York Telephone, the number 7, represents A. interest or coupon rate.

The bond interest rate is the annual interest (coupon) rate paid on a bond.  

The amount of interest is expressed as a percentage of the face value.

The bond interest (rate or amount) represents the periodic compensation that is due in exchange for the bondholder's agreement to lend their funds to the entity for some period.

Thus, the number 7, represents A. interest or coupon rate., not the maturity date, current yield, or the difference in price.

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Answer: interest or coupon rate

Step-by-step explanation:

got 100% on the test

Hope this helps a little

Answer:

Ryan sprinted 46.33 yards.

Step-by-step explanation:

Ryan sprinted 139 feet.

1 feet = 0.33 Yard

Formula: Divide the length value by 3

Now,

139 ÷ 3 = 46.33

Thus, Ryan sprinted 46.33 yards.

-TheUnknownScientist

\(f(x) = 2x is linear.f(x) = 4^x is exponential.\) This is where at each level, each branch extends into 4 branches.

In the first prompt, the number of boxes in each row is increasing by a constant amount, so the function is linear. In the second prompt, the number of branches is increasing by a factor of 4 with each level, so the function is exponential.

A linear function is defined as one where the output increases by a constant amount for each increase in the input. An exponential function is defined as one where the output increases by a constant factor for each increase in the input. The first prompt corresponds to a linear function, while the second prompt corresponds to an exponential function.

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The complete question is-

Is the function f(x) = 4^x, where x is the number of branches at level x in a tree diagram, linear or exponential? Explain your answer.

1.Bricks and timber purchased for construction. (Debit: Bricks - Rs 60,000, Debit: Timber - Rs 35,000, Credit: Bank - Rs 95,000)

2.Transfer of Rs 13,000 to fixed deposit account. (Debit: Fixed Deposit - Rs 13,000, Credit: Current Account - Rs 13,000)

3.Appointment of Mr. S.N. Rao as Accountant. (Debit: Salary Expense - Rs 300, Debit: Security Deposit - Rs 1,000, Credit: Accountant - Rs 300)

4.Goods sold to Shruti with discounts. (Debit: Accounts Receivable - Shruti - Rs 80,000, Credit: Sales - Rs 80,000)

5.Goods purchased from Richa with discounts. (Debit: Purchases - Rs 60,000, Credit: Accounts Payable - Richa - Rs 60,000)

6.Goods sold to Shilpa with discounts and received payment. (Debit: Accounts Receivable - Shilpa - Rs 50,000, Credit: Sales - Rs 50,000)

7.Goods sold to Garima with discounts and received partial payment. (Debit: Accounts Receivable - Garima - Rs 1,00,000, Credit: Sales - Rs 1,00,000)

8.Purchase of land with additional charges. (Debit: Land - Rs 2,00,000, Debit: Brokerage Expense - Rs 2,000, Debit: Registration Charges - Rs 15,000, Credit: Bank - Rs 2,17,000)

9.Proprietor took goods and cash for personal use. (Debit: Proprietor's Drawings - Rs 65,000, Credit: Goods - Rs 25,000, Credit: Cash - Rs 40,000)

10.Goods sold to Charu with profit and discount. (Debit: Accounts Receivable - Charu - Rs 1,20,000, Credit: Sales - Rs 1,20,000)

11.Rent paid for the building. (Debit: Rent Expense - Rs 60,000, Credit: Bank - Rs 60,000)

12.Goods sold to Sunil with profit and discount. (Debit: Accounts Receivable - Sunil - Rs 24,000, Credit: Sales - Rs 24,000)

13.Purchased goods from Nanda and supplied to Helen. (Debit: Purchases - Rs 1,000, Debit: Accounts Payable - Nanda - Rs 1,000, Credit: Accounts Receivable - Helen - Rs 1,300, Credit: Sales - Rs 1,300)

14.Purchased goods from Rohit and Sons and supplied to Madan. (Debit: Purchases - Rs 2,700, Credit: Accounts Payable - Rohit and Sons - Rs 2,700, Debit: Accounts Receivable - Madan - Rs 3,000, Credit: Sales - Rs 3,000)

Here are the journal entries for the given transactions:

1. Bricks and timber purchased for construction:

Debit: Bricks (Asset) - Rs 60,000

Debit: Timber (Asset) - Rs 35,000

Credit: Bank (Liability) - Rs 95,000

2. Transfer to fixed deposit account:

Debit: Fixed Deposit (Asset) - Rs 13,000

Credit: Current Account (Asset) - Rs 13,000

3. Appointment of Mr. S.N. Rao as Accountant:

Debit: Salary Expense (Expense) - Rs 300

Debit: Security Deposit (Asset) - Rs 1,000

Credit: Accountant (Liability) - Rs 300

4. Goods sold to Shruti:

Debit: Accounts Receivable - Shruti (Asset) - Rs 80,000

Credit: Sales (Income) - Rs 80,000

5. Goods purchased from Richa:

Debit: Purchases (Expense) - Rs 60,000

Credit: Accounts Payable - Richa (Liability) - Rs 60,000

6. Goods sold to Shilpa:

Debit: Accounts Receivable - Shilpa (Asset) - Rs 50,000

Credit: Sales (Income) - Rs 50,000

7. Goods sold to Garima:

Debit: Accounts Receivable - Garima (Asset) - Rs 1,00,000

Credit: Sales (Income) - Rs 1,00,000

8.Purchase of land:

Debit: Land (Asset) - Rs 2,00,000

Debit: Brokerage Expense (Expense) - Rs 2,000

Debit: Registration Charges (Expense) - Rs 15,000

Credit: Bank (Liability) - Rs 2,17,000

9. Goods and cash taken away by proprietor:

Debit: Proprietor's Drawings (Equity) - Rs 65,000

Credit: Goods (Asset) - Rs 25,000

Credit: Cash (Asset) - Rs 40,000

10. Goods sold to Charu:

Debit: Accounts Receivable - Charu (Asset) - Rs 1,20,000

Credit: Sales (Income) - Rs 1,20,000

Credit: Cost of Goods Sold (Expense) - Rs 80,000

Credit: Profit on Sales (Income) - Rs 40,000

11. Rent paid for the building:

Debit: Rent Expense (Expense) - Rs 60,000

Credit: Bank (Liability) - Rs 60,000

12. Goods sold to Sunil:

Debit: Accounts Receivable - Sunil (Asset) - Rs 24,000

Credit: Sales (Income) - Rs 24,000

Credit: Cost of Goods Sold (Expense) - Rs 20,000

Credit: Profit on Sales (Income) - Rs 4,000

13. Goods purchased from Nanda and supplied to Helen:

Debit: Purchases (Expense) - Rs 1,000

Debit: Accounts Payable - Nanda (Liability) - Rs 1,000

Credit: Accounts Receivable - Helen (Asset) - Rs 1,300

Credit: Sales (Income) - Rs 1,300

14. Goods received from Rohit and Sons and supplied to Madan:

Debit: Purchases (Expense) - Rs 2,700 (after 10% trade discount)

Credit: Accounts Payable - Rohit and Sons (Liability) - Rs 2,700

Debit: Accounts Receivable - Madan (Asset) - Rs 3,000

Credit: Sales (Income) - Rs 3,000

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Answer:1

Step-by-step explanation: so bro if yo do 5*1/25 u can see

Answer: 61.44

Step-by-step explanation:

To find the quotient of 9216 and 150, it means to divide.

9216/150=61.44

Rounding to two decimal places, the Z-score corresponding to the youngest senator of age 39 is -1.40.

To find the Z-score corresponding to a particular value in a normal distribution, we use the formula:

Z = (x - μ) / σ

where x is the value of interest, μ is the mean, and σ is the standard deviation.

Part 1 of 2: For the oldest senator of age 85:

Z = (x - μ) / σ = (85 - 56.5) / 12.5 ≈ 2.28

Rounding to two decimal places, the Z-score corresponding to the oldest senator of age 85 is 2.28.

Part 2 of 2: For the youngest senator of age 39:

Z = (x - μ) / σ = (39 - 56.5) / 12.5 ≈ -1.4

Rounding to two decimal places, the Z-score corresponding to the youngest senator of age 39 is -1.40.

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Ab=3.14xRadious to the power of 2

then to the area of the rectange you do B x H = YOUR ANSWER

The dimensions of the smaller rectangle are 13 centimeters by 52 centimeters.

Let's assume the dimensions of the smaller rectangle are length L and width W (in centimeters).

We know that the area of the smaller rectangle is 676 square centimeters:

L * W = 676    ----(1)

We also know that the larger rectangle has a shorter side of 41 centimeters. Let's say the corresponding longer side of the larger rectangle is H centimeters.

The area of the larger rectangle is 3457 square centimeters:

41 * H = 3457    ----(2)

Our current set of equations contains two unknowns. In order to get the smaller rectangle's dimensions, we can simultaneously solve these equations.

In order to find H, we can use equation (2):

H = 3457 / 41

H ≈ 84.22

Now we can substitute this value of H into equation (1):

L * W = 676

We need to find the dimensions (L and W) that multiply to give 676. We can start by looking for factors of 676.

Factors of 676: 1, 2, 4, 13, 26, 52, 169, 338, 676

By trial and error, we can see that the factors that give a close match to the dimensions of the larger rectangle (41 and 84.22) are 13 and 52:

L = 13

W = 52

Therefore, the smaller rectangle has measurements of 13 by 52 centimetres.

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Answer:

-9/14 so the answer is B

Step-by-step explanation:

Hope this helps

Answer:

b <3

Step-by-step explanation:

edge 2021

Answer:

\(y^{\frac{3}{5}} =\sqrt[5]{y^{3} }\)

Step-by-step explanation:

\(y^{\frac{3}{5}} =\sqrt[5]{y^{3} }\)

Answer:

ewannnnnnnnnnnnnn

Step-by-step explanation:

d q den alam d pa namin pinagaaralan yan, xenxia na godbless