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2 years ago

10

Write a two-column proof for the following information. Upload your proof below.

1 answer:

Evgesh-ka [11]2 years ago

4 0

Statements Reasons

1.) M is the midpoint of CD; CM = 5x – 2; MD = 3x + 2 1.) Given

2.) CM=MD 2.) Definition of midpoint

3.) 5x-2=3x+2 3.) Substitution

4.) 2x-2=2 4.) Subtraction Property of Equality

5.) 2x=4 5.) Subtraction Property of Equality

6.) x=2 6.) Division Property of Equality

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Please anyone answer me

ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

$A\frac{}{1} = ∫g(x)dx = ∫xdx = \frac{1}{2} {x}^{2} | \frac{2}{0} = 2$

Area 2:

$A\frac{}{2} = ∫(g(x) - f(x))dx$

$= ∫(x - {(x - 2)}^{2} )dx$

$= ∫( - {x}^{2} + 5x - 4)dx$

$= ( - \frac{1}{3}{x}^{3} + \frac{5}{2} {x}^{2} - 4x) | \frac{4}{2}$

$= 2.67 - ( - 0.67) = 3.34$

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0

3 years ago

Erin received the following scores on her last four spelling tests.

Tatiana [17]

The second one, 82% is Erin's highest score

4 0

3 years ago

Read 2 more answers

Marcel said he can find 52 times 10 by adding a 0 to the end of 52 to get 520

Rainbow [258]

Answer:

No he can not because if you add 0 to 52 you get 520 and the answer has to be 520

Step-by-step explanation:

4 0

3 years ago

the length of a rectangle is 4cm longer than the width. the perimeter is 80cm. find the length and the width.

skelet666 [1.2K]

Answer:

I think it is 20cm

Step-by-step explanation:

divide 80 by 4 =20 cm

7 0

3 years ago

Charlie's garage has a rectangular floor space. Its length is 3 times its width. Charlie extends the width of his garage to 6 m,

MAXImum [283]

Answer:

The percentage change is 140%

Step-by-step explanation:

Given

$L= 3W_1$ ---- initial dimension

$W_2 = 6m$ --- new width

$A_2 = 45m^2$ --- new dimension

Required

The percentage increment

The length remains constant because only the width is extended.

The new area is:

$Area =Length * Width$

$A_2=L * W_2$

Make L the subject

$L = \frac{A_2}{ W_2}$

Substitute values for A and W

$L = \frac{45m^2}{6m}$

$L = \frac{45m}{6}$

$L = 7.5m$ --- this is the length of the garden

Calculate the initial width:

$L= 3W_1$

Make W1 the subject

$W_1 = \frac{1}{3} * L$

$W_1 = \frac{1}{3} * 7.5$

$W_1 = 2.5$

So, the initial area is:

$A_1 = L_1 * W_1$

$A_1 = 2.5 * 7.5$

$A_1 = 18.75$

The percentage change in area is:

$\%A = \frac{A_2 - A_1}{A_1}$

$\%A = \frac{45 - 18.75}{18.75}$

$\%A = \frac{26.25}{18.75}$

$\%A = 1.4$

Express as percentage

$\%A = 1.4*100\%$

$\%A = 140\%$

4 0

3 years ago