a. Write 60,000,000 in words.

Answer:

C. .024

Step-by-step explanation:

You multiply 25 by 4 to get 100 and then you multiply 6 by 4 to get twenty four. After that you divide both numbers by one hundred and you get C

Answer:

47.53

Step-by-step explanation:

First you add all of the numbers together.

66.9+5.6+70.1=142.6

Then you divide the sum of numbers by how many number there are.

142.6/3=47.53

Answer:

i believe it's 39

Step-by-step explanation:

29 -39 = -10

Answer:

the answer is 2.16

Step-by-step explanation:

just use a calculator and it will tell you the answer

Answer:

4

Step-by-step explanation:

Answer:

a.  y= e raise to power y

c. y = e^ky

Step-by-step explanation:

The first derivative is obtained by making the exponent the coefficient and decreasing the exponent by 1 . In simple form the first derivative of

x³ would be 2x³-² or 2x².

But when we take the first derivative of  y= e raise to power y

we get y= e raise to power y. This is because the derivative of e raise to power is equal to e raise to power y.

On simplification

y= e^y

Applying ln to both sides

lny= ln (e^y)

lny= 1

Now we can apply chain rule to solve ln of y

lny = 1

1/y y~= 1

y`= y

therefore

derivative of e^y = e^y

The chain rule states that when we have a function having one variable and one exponent then we first take the derivative w.r.t to the exponent and then with respect to the function.

Similarly when we take the first derivative of  y= e raise to power ky

we get y=k multiplied with e raise to power ky. This is because the derivative of e raise to a  constant and power is equal to constant multiplied  with e raise to power y.

On simplification

y= k e^ky

Applying ln to both sides

lny=k ln (e^y)

lny=ln k

Now we can apply chain rule to solve ln of y ( ln of constant would give a constant)

lny = ln k

1/y y~= k

y`=k y

therefore

derivative of e^ky = ke^ky

Answer:

3y-2

Step-by-step explanation:

y+y+y-2

3y-2

The perimeter of the isosceles triangle with base length y - 2 and length of legs y is 3y - 2. legs.

Here,

Base length of triangle is y - 2.

Length of legs of triangle is y.

What is Perimeter of triangle?

Perimeter of triangle is sum of all three sides of a triangle.

Now,

Base length of triangle is y - 2.

Length of legs of triangle is y.

Perimeter = y - 2 + y + y

               = 3y - 2

Hence, The perimeter of the isosceles triangle with base length y - 2 and legs of length y is 3y - 2.

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Answer:

Step-by-step explanation:

Answer:

3m ≤ ∫ f(x) dx ≤ 3M, at limit of b, a

Step-by-step explanation:

Like the question asked, which property of integral was used.

Property 8 of integrals was the basis upon which the question was solved.

The property of the integral is used to solve the problem. The function is given below.

[tex]3m & \leq \int_a^b f(x)dx \leq 3M \end{aligned}[/tex]

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

Suppose f has absolute minimum value m and absolute maximum value M.

We know that the property of the integral can be used.

If m ≤ f(x) ≤ M ∀ a ≤ x ≤ b. Then we have

[tex]m (b-a) \leq \int_a^b f(x) dx \leq M(b-a)\\[/tex]

Then we have

[tex]m \leq f(x) \leq M \ \ and \ \ 4 \leq x \leq 7[/tex]

This means that

[tex]\begin{aligned} m(7-4) & \leq \int_a^b f(x)dx \leq M(7-4)\\\\3m & \leq \int_a^b f(x)dx \leq 3M \end{aligned}[/tex]

More about the function link is given below.

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w²+7w>293 is the inequality which we use to determine what lengths would make the area of the base of the tree house greater than 293 square Inches

The area of Rectangle is length times of width.

Given that Adam built a tree house with a rectangular base.

The length of the base is 7 inches more than its width.

Length=W+7

W is the width

The area of the base of the tree house greater than 293

We need to find the inequality to represent the area of the base of tree

Area of rectangle=Length×width

293=(w+7)w

293=w²+7w

As area is greater we have

w²+7w>293

Hence, w²+7w>293 is the inequality which we use to determine what lengths would make the area of the base of the tree house greater than 293 square Inches

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Answer:

Step-by-step explanation:

ans is 25

Step-by-step explanation:

12(10-5)-40+(4+1)

=12 of 5 - 40 + 5

= 60 - 40 +5

=60 +5 - 40

=65-40

=25

Answer:

-25

Step-by-step explanation:

We start off by using PEMDAS. 10-5 then 4+1.

12(5)-40+5

we multiply, then add and subtract giving us -25

Answer:

Here Ratio=1:2:4

total money =$70

Step-by-step explanation:

let the ratio number be 1x,2x and4x

Now,

1x+2x+4x =$70

or, 7x. =$70

or, x. =$70÷7=$10

again,

1x=1×$10=$10,

2x=2×$10=$20 and

4x=4×$10=$40

Answer:

18

Step-by-step explanation:

The integral is the area between f(x) and the x-axis.  Remember that area below the x-axis is negative.

The area between x=0 and x=6 is a trapezoid.

The area between x=6 and x=9 is a rectangle.

The area between x=9 and x=15 is a triangle.

The area between x=15 and x=21 is a triangle.

The area between x=21 and x=27 is a trapezoid.

A = ½ (3+9) (6) + (3)(9) + ½ (6)(9) + ½ (6)(-9) + ½ (-9+-6) (6)

A = 36 + 27 + 27 − 27 − 45

A = 18

Answer:

Step-by-step explanation:

3x^2 - 4x^2 - 6mn + 5mn

-x^2 - mn

Answer:

AB = 4.5

Step-by-step explanation:

Distance between A(-1, 2) and B(1, -2):

[tex] AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]

Let,

[tex] A(-1, 2) = (x_1, y_1) [/tex]

[tex] B(1, -2) = (x_2, y_2) [/tex]

[tex] AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2} [/tex]

[tex] AB = \sqrt{(2)^2 + (-4)^2} [/tex]

[tex] AB = \sqrt{4 + 16} = \sqrt{20} [/tex]

[tex] AB = 4.5 [/tex] (nearest tenth)

Answer:

The saying money breeds money means that with money, you have the possibility to create more money. Simple interest is the idea that you have something, and you obtain a percentage of that every designated time period. That being said, simple interest = money breeds money

Answer: Area of Polygon Q is  [tex]\dfrac14[/tex]  of  Area of Polygon

Step-by-step explanation:

Given: Polygon Q is a scaled copy of Polygon P using a scale factor of [tex]\dfrac12[/tex] .

According to the property of scale factor ,

Area of the image = (scale factor )²x (Area of the original figure)

So, Area of Polygon Q = [tex](\dfrac12)^2[/tex]  x (Area of Polygon P)

⇒ Area of Polygon Q = [tex]\dfrac14[/tex]  x (Area of Polygon P)

Hence,  Area of Polygon Q is   [tex]\dfrac14[/tex]  of  Area of Polygon P.

Answer:

1/4

I did on khan

Step-by-step explanation:

Answer:

12, 12 and 12 so 36

Step-by-step explanation:

Answer:

your pics a little blurry but if i am reading it right

the first one

f(x)=log x+5

Answer:

[tex]\frac{x}{5} + 6y[/tex]

Step-by-step explanation:

Answer:

The surface area of the larger prism is 4 times the surface area of the smaller one.

(which agrees with answer B in your list)

Step-by-step explanation:

Notice that the lateral surface area of prisms always consists of parallelograms, whose areas are given by the product base times height. = B x H

Therefore if the dimensions of the prism double, then the base and height of the parallelograms will double as well, making such individual ares of the lateral faces go from:

B x H  to -->  2 B x 2 H = 4 (B x H)

that is 4 times the original lateral face's area.

with the faces for the top and bottom bases of the prism, something similar happens, so we conclude that the surface area of the larger prism is 4 times the surface area of the smaller one.

Answer:

c

Step-by-step explanation:

edge 2021

Answer:

Seth

Step-by-step explanation:

3/12 = 0.25

0.25*100 = 25%

Seth scored at 25%

2/10 = 0.2

0.2*100 = 0.2

0.2*100 = 20%

Zak scored at 20%

then:

25>20

then:

Seth scored on a greater fraction of his shots.

Zak scored on a greater fraction of his shots.

Given,

In a hockey game, Seth took 12 shots and scored 3 times.

Zak took 10 shots and scored twice.

We need to find who scored on a greater fraction of his shots.

We can compare fractions by finding their decimal values a nd comparing them.

Example: 1/2 and 2/5

1/2 = 0.5

2/5 = 0.4

0.5 is greater than 0.4.

Find the number of score Seth shots out of 12 shots.

= 3

We can write in fractions as:

3/12 _____(1)

Find the number of score Zak shots out of 10 shots.

= 2 x 3

= 6

We can write in fractions as:

6/10 _____(2)

Compare (1) and (2)

3/12 = 1/4 = 0.25
6/10 = 3/5 = 0.6

0.25 is less than 0.6

6/10 is a greater fraction than 3/12

Thus Zak scored on a greater fraction of his shots.

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step 1: open up the brackets ( rmb to flip the signs )

step 2: bring 'a' to one side and the numbers to the other

6a + 34 = -3(-2a + 8)

6a + 34 = 6a - 24

6a - 6a = -24 - 34

0 = - 58

the final ans is kind of weird but here's my solution :))

Answer:

Step-by-step explanation:

If the distance marked on a centimeter scale is from 2.5 to 4.5, then the final destination is 4.5cm and the initial point = 2.5.

Distance between the two points will give us the length of the line segment.

Distance between the two points = final point - initial point

Distance between the two points  = 4.5 - 2.5

Distance between the two points = 2.0 cm

Hence the line segment is 2.0cm long

Answer:

(a) 0.16759

(b)  0.9649

Step-by-step explanation:

Given that:

n = 10 , p = 0.3 and  [tex]\hat p = 0.4[/tex]

[tex]P(|\hat p - p| > 0.2 ) = 1 - P ( |\hat p -p| \leq 0.2)[/tex]

= [tex]1 - P(-0.2 \leq \hat p-p \leq 0.2)[/tex]

= [tex]1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{\hat p -p}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{0.2}{\sqrt{\dfrac{pq}{n}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{0.021}} \leq Z \leq \dfrac{0.2}{\sqrt{0.021}} \end {pmatrix}[/tex]

[tex]=1 - P \begin {pmatrix} -1.380 \leq Z \leq 1.380 \end {pmatrix}[/tex]

= 1 - P( Z ≤ 1.380) - P(-1.380)

= 1 - ( 0.91620 - 0.08379 )

= 1 - 0.83241

= 0.16759

b) when n = 400; p =0.3 , q = 1 - p = 1 - 0.3 = 0.7

[tex]P( |\hat p - p | > 0.001) = 1- P ( |\hat p - p | < 0.001 )[/tex]

[tex]= 1- P ( -0.001 < \hat p - p < 0.001 )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{pq}{n}}} < \dfrac{ \hat p - p}{\dfrac{pq}{n}} < \dfrac{0.001}{\sqrt{\dfrac{pq}{n}}} )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{0.3\times 0.7}{400}}} < Z < \dfrac{0.001}{\sqrt{\dfrac{0.3 \times 0.7}{400}}} )[/tex]

[tex]= 1- P ( \dfrac{-0.001}{\sqrt{5.25 \times 10^{-4}}} < Z < \dfrac{0.001}{\sqrt{5.25 \times 10^{-4}}} )[/tex]

[tex]= 1- P ( -0.0436< Z < 0.0436)[/tex]

= 1 - P ( Z < 0.0436) - P ( -0.0436)

= 1 - (0.5176 - 0.4825)

= 1 -  0.0351

= 0.9649

Answer:

B) -2

Step-by-step explanation:

[tex]\frac{25-(-15)}{-12-8} =\frac{40}{-20} =-2[/tex]