Can someone help me with the questions in the picture?

Answer:

last one

Step-by-step explanation:

Answer:

4 (6-k)

the bottom one

Step-by-step explanation:

Answer:

length = 20 feet

Step-by-step explanation:

the area (A) of a rectangle is calculated as

A = length × width

let width be w , then length is 4w

given A = 100 , then

4w × w = 100

4w² = 100 ( divide both sides by 4 )

w² = 25 ( take square root of both sides )

\(\sqrt{w^2}\) = \(\sqrt{25}\)

w = 5

width = 5 feet and length = 4 × 5 = 20 feet

Answer:

8x - 4

Step-by-step explanation:

Perimeter is the sun of all side lengths. We know the perimeter and two side lengths, so on order to find the third, we need to subtract the first two lengths from the sum. After subtraction, we get 8x - 4.

Answer:

e

Step-by-step explanation:

a

The result of the division of dividend 3 1/6 by divisor 1 3/8 is computed as 76/33.

For the stated question;

The dividend is  3 1/6, which is given in mixed fraction.

In proper fraction; 3 1/6 = 19/6.

The divisor is  1 3/8 = 11/8.

Thus,

= 19/6 / 11/8

= 19*8 / 6*11

= 76/33

Thus, the result of the division of dividend 3 1/6 by divisor 1 3/8 is computed as 76/33.

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Answer:

c = 17 ft

Step-by-step explanation:

c=a2+b2=82+152=17ft

Answer:

17

Step-by-step explanation:

a2+b2=c2

8^2+15^2=c2

64 + 225 = 289

√289 = 17

hope it helps :-)

Answer:

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

Find the hypotenuse c of the base using Pythagorean theorem:

Lateral surface area, three rectangular faces, is:

Find base area, the area of two right triangles:

Find total surface area:

Answer:

Step-by-step explanation:

7

Answer:

7

Step-by-step explanation:

3.5 divided by .5

Answer:

Q'(-2 , 5)

Step-by-step explanation:

Note that:

1) When you are reflecting over the x-axis, you are changing the sign for the y.

2) When you are reflecting over the y-axis, you are changing the sign for the x.

In this case, you are reflecting over the y-axis. Change the sign of the x:

Q(2 , 5) ⇒ Q'(-2 , 5)

Q'(-2 , 5) is your answer.

~

The approximate ordered pair for f(x) = g(x) is given as; (4, 22) from the graph. Option A

We know that there are two equations that can be shown from the set of equations that have been written here and the set of equations that we have are;

f(x) = −2x + 30

g(x) = 3x + 10

We are now trying to see the points at which the equation can be said to be equal. This refers to the point where  f(x) = g(x). How can we be able to obtain this point? If we are to get this point then we must go back and look at the graph very well.

As we are looking at the graph very well we should fix our eyes on any point where the line of two functions can be seen to intersect. Once that happens, we can say that we have reached the point where the two functions can be said to be equal. This point is read from the graph as (4, 22)

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Answer:

They have 534 cards left

Step-by-step explanation:

The closed shapes with four sides are called quadrilaterals.

Therefore, the four sides closed shape is known as quadrilateral.

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Answer: 90

Step-by-step explanation:

Bracket

Indices

Division

Multiplication

Addition

Subtraction

I: \(5^{2}\) = 5 x 5 = 25

M: 3 x 25 = 75

A: 15 + 75 = 90

Answer:

1. For a rotation of 270° counterclockwise about the origin the coordinates of J' is (1, -2)

2. For a rotation of 180° counterclockwise about the origin the coordinates of K' is (-7, -1)

3. For a rotation of 90° counterclockwise about the origin the coordinates of L' is (3, 6)

4. For a rotation of 90° counterclockwise about the origin the coordinates of M' is (3, 1)

Step-by-step explanation:

The coordinates of the vertices of the parallelogram are given as follows;

J(2, 1), K(7, 1), L(6, -3), M(1, -3)

1. We have, for a rotation of 270° counterclockwise about the origin we have

f(x, y) = (y, -x)

Therefore, we have for the vertices;

f(J(2, 1)) = J'(1, -2)

Therefore, we have the coordinates of J' as (1, -2);

2. We have, for a rotation of 180° counterclockwise about the origin we have

f(x, y) = (-x, -y)

Therefore, we have for the vertex point K;

f(K(7, 1)) = K'(-7, -1)

3. We have, for a rotation of 90° counterclockwise about the origin we have

f(x, y) = (-y, x)

Therefore, we have for the vertex point L;

f(L(6, -3)) = L'(3, 6)

4. We have, for a rotation of 90° counterclockwise about the origin we have

f(x, y) = (-y, x)

Therefore, we have for the vertex point M;

f(M(1, -3)) = M'(3, 1)

Answer:

Step-by-step explanation:

One pair of coordinates is sufficient to work out the constant of proportionality.

Equation in general form:

and

Substitute x- and y- coordinates into equation and find the value of k:

Answer:

-x² + 13x + 2

Step-by-step explanation:

you are on ambitious level, and you cannot answer that yourself ? that is trivial. truly. no hidden traps, just straight forward adding and simplifying. what is the problem, please ?

-7x² + 6x² = -x²

5x + 8x = 13x

-1 + 3 = 2

Answer:

3rd option

Step-by-step explanation:

Given

(- 7x² + 5x - 1) + (6x² + 8x + 3) ← remove parenthesis

= - 7x² + 5x - 1  + 6x² + 8x + 3 ← collect like terms

= - x² + 13x + 2

Answer:

Here are the completed ratio tables for both classes:

Mrs. Washington's class:

Band: 2

Chorus: 3

Total: 2 + 3 = 5

Band-to-Chorus ratio: 2 : 3

Mr. Utley's class:

Band: 5

Chorus: 7

Total: 5 + 7 = 12

Band-to-Chorus ratio: 5 : 7

Based on the ratios, it appears that Mr. Utley's class has a greater ratio of students in band to students in chorus. The ratio in Mr. Utley's class is 5 : 7, while the ratio in Mrs. Washington's class is 2 : 3.

Step-by-step explanation:

Step-by-step explanation:

radius are a special form of fractions. but they are fractions, and a such all rules of calculation and comparison apply.

mrs. Washington's class has a ratio of 2/3 (or 2:3 or 2÷3).

2/3 = 0.666666666...

mrs. utley's class has a ratio of 5/7.

5/7 = 0.714285714...

so, the class of mrs. Utley had a greater ratio.

to compare the ratios (like any fraction) within calculating their decimal value, we need to bring them to the same denominator :

.../3 and .../7

we need to bring both to .../21 :

2/3 × 7/7 = 14/21

5/7 × 3/3 = 15/21

so, we can clearly see that the ratio of mrs. Utley's class is greater, as 15 is greater than 14.

Answer:

95%

Step-by-step explanation:

In this question, it's important to note that the question states the distribution is a normal distribution, therefore we can use the Empirical Rule (68-95-99.7 rule).

The Empirical Rule, also known as the 68-95-99.7 rule, states that:

*Note that the Empirical Rule only works for normal distributions

We know that the mean, \(\mu\), is given as 79 and the standard deviation, \(\sigma\), is given as 17. Since the standard deviation is 17, 2 standard deviations must be \(2\sigma=2\cdot 17=34\) (and 3 standard deviations would just be 51).

We want to find the percentage of scores (% of data) that falls between the scores 45 and 113.

Notice that 45 is 34 less than the mean of 79 and 113 is 34 more than the mean of 79, therefore, the range of scores between 45 and 113 varies between ±2 standard deviations from the mean.

From the Empirical Rule, we can see that 95% of data falls between ±2 standard deviations in a normal distributions.

Thus, 95% of scores fall between 45 and 113.

Using the given exponential function,

The rate of the disease spreading is approx. 140 peoples per day affected.

let P be the number of people's that are contacting a new disease.

given exponential function about the new disease modeled by population is

P(t) = 3725/(1 +1.72 e⁻⁰·⁵²ᵗ) ---(1)

where t is time measured in days

we want to calculate the rate of diseases spreading on day 4 i.e t = 4

putting the value t = 4 in above formula, we get

P(t=4) = 3725/(1 + 1.72 e⁻⁽⁰·⁵²⁾⁴)

=> P( t= 4) = 3725 / (1 + 1.72× e⁻⁰·²⁸)

=> P(t=4) = 3725/(1 + 1.72× 0.124)

=> P(t = 4) = 3725/(1+ 0.2148)

=> P(t=4) = 3725/1.2148

=> P(t = 4) = 3066.14

Hence, after 4 days of diseases spread , total 3066 people's are affected from it .

rate of diseases spreading per day is equal to (3725-3066)/4 = 659/4 = 139.75

so, rate of spreding diseases is 139.75 ~ 140 peoples per day .

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Verifying that this function satisfies the three hypotheses of Rolle's Theorem on the inverval f(x) is f(x) is and f(0) on [0, 4] on (0, 4) f(4) Then by Rolle's theorem, there exists at least one value c = 2  such that f'(c) = 0.

Given:

Consider the function f(x) = x2 - 4x + 8 on the interval 0, 4. Verify that this function satisfies the three hypotheses of Rolle's Theorem on the inverval f(x) is f(x) is and f(0) on [0, 4] on (0, 4) f(4) Then by Rolle's theorem, there exists at least one value c such that f'(c) = 0.

f(x)=x^2−4x+8, [0,4]

when, x = 0

f(x) = x^2 -4x +8

f(0) = y = 0 - 0 + 8 = 8

when, x=4

f(5) = y = 16 - 16+8 =

thus, we have 2 points (0, 8) ; (4, 8)

slope,m = {8-(8)} / {4-0} = 0

hence, we have to calculate all the points,x where 0<x<8 and slope=0

f '(x) = 2x - 4 = 0

or, f '(c) = 2c - 4 = 0

c = 4/2 =2 ( 0<x<4)

hence, the there is only one solution c=2 which satisfies Rolle's theorem.

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Answer:

x > -39

Step-by-step explanation:

-0.3x < 6.7 + 5

-0.3x < 11.7

divide by -0.3

Answer is x > -39

To evaluate the second derivative of the function f(x) = (x - 9)/(5x + 6) at the points x = 3, x = 0, and x = -2, we first need to find the first derivative and then  the second derivative.  And the second derivative f''(x) of the function f(x) = (x - 9)/(5x + 6) is constantly equal to 0

Step 1: Finding the first derivative:

To find the first derivative f'(x), we apply the quotient rule. Let's denote f(x) as u(x)/v(x), where u(x) = x - 9 and v(x) = 5x + 6. Then the quotient rule states:

f'(x) = (u'(x)v(x) - v'(x)u(x))/(v(x))^2

Applying the quotient rule, we get:

f'(x) = [(1)(5x + 6) - (5)(x - 9)]/[(5x + 6)^2]

      = (5x + 6 - 5x + 45)/[(5x + 6)^2]

      = 51/[(5x + 6)^2]

Step 2: Finding the second derivative:

To find the second derivative f''(x), we differentiate f'(x) with respect to x:

f''(x) = [d/dx(51)]/[(5x + 6)^2]

       = 0/[(5x + 6)^2]

       = 0

The second derivative f''(x) is a constant value of 0, which means it does not depend on the value of x. Therefore, the second derivative is constant and does not change with different values of x.

Now, let's evaluate f''(3), f''(0), and f''(-2):

f''(3) = 0

f''(0) = 0

f''(-2) = 0

In summary, the second derivative f''(x) of the function f(x) = (x - 9)/(5x + 6) is constantly equal to 0 for any value of x. Hence, f''(3), f''(0), and f''(-2) all evaluate to 0.

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Answer:

-60+30=(-30)

Step-by-step explanation: The whale was swimming 60 feet BELOW sea level. That's negative. Then it swims up 30 feet, that's positive. Add to get -30, since -60 is the greater number although not in value. you get -30, so move the dot up , and the whale is now 30 feet below the surface.

Please let me know if this helped.

Claire fed Folly at 2 pm, gave Folly to Max at 2:10 pm. Folly belonged to either Max or Claire at 2:05 pm. Neither Mar nor Claire fed Folly at 2 pm or 2:05 pm. The event occurred between 1:55 pm and 2:05 pm. At 2 pm, Max took Folly from Claire. Folly wasn't hungry at 2 pm but was at 3 pm.

1. student(Mar) ∧ ¬pet(Mar)2. fed(Claire, Folly, 2pm) ∧ gave(Claire, Folly, Max, 2:10pm)3. (belongs(Folly, Max, 2:05pm) ∨ belongs(Folly, Claire, 2:05pm))4. ¬(fed(Mar, Folly, 2pm) ∨ fed(Claire, Folly, 2pm) ∨ fed(Mar, Folly, 2:05pm) ∨ fed(Claire, Folly, 2:05pm))5. between(2pm, 1:55pm, 2:05pm)6. ¬hungry(Folly, 2pm) ∧ hourLater(Folly, 2pm, 3pm)

1. Student(Mar) ∧ ¬Pet(Mar)2. Fed(Claire, Folly, 2pm) ∧ Gave(Claire, Folly, Max, 2:10pm)3. BelongsTo(Folly, Max, 2:05pm) ∨ BelongsTo(Folly, Claire, 2:05pm)4. ¬(Fed(Mar, Folly, 2pm) ∨ Fed(Claire, Folly, 2pm) ∨ Fed(Mar, Folly, 2:05pm) ∨ Fed(Claire, Folly, 2:05pm))

5. Between(2:00pm, 1:55pm, 2:05pm)6. Gave(Max, Folly, Claire, 2pm) ∧ ¬Hungry(Folly, 2pm) ∧ Hungry(Folly, 3pm)

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Answer:

9:20

Step-by-step explanation:

9 males

11 + 9 total people

9 : 20

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

Answer:

Yes

Step-by-step explanation:

Yes because if you insert -39 for x and 42 for y then your equation is:

42 = -39 + 81

If you were to solve this equation it would be true.

The statements that is true about the similarity of the two triangles the option D

D. ΔMNO and ΔJKL are not similar triangles

Similar triangles are triangles which have proportional corresponding sides

The parameters in the question are;

The length of segment MN = 20

Length of segment NO = 12

Length of segment OM = 25

Measure of angle ∠O = 56°

Length of segment LJ =- 15

Length of segment JK = 12

Length of segment KL = 9

Measure of angle ∠L = 56°

Two triangles are similar if the ratio of two sides on one triangle are proportional to two sides of another triangle, and the included  angle between the two sides are congruent

The included angle between sides ON and MO on triangle MNO is congruent to the included angle between segment LK and JL in triangle JKL

However, the ratio of the sides  LK to ON and JL to MO are;

9/12 ≠ 15/25

Therefore, the triangles are not similar

The correct option is option D

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