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3 years ago

6

Is my answer correct?

1 answer:

Elenna [48]3 years ago

5 0

Answer:

you are right

Step-by-step explanation:

You might be interested in

How are the solutions to the following two equations related m+5=72 <br> m+72=5

EastWind [94]

To see how they are related, let us solve the two equations.

FIRST EQUATION

$m + 5 = 72$

$\rightarrow \: m = 72 - 5$

$\rightarrow \: m = 67$

SECOND EQUATION

$m + 72 = 5$

$\rightarrow \: m = 5 - 72$

$\rightarrow \: m = - 67$

Hence the solutions to the two equations are ADDITIVE INVERSE of each other.

6 0

4 years ago

Help pls it easy for everyone, i need it before 20 mins plss

Katen [24]

5.B
6.D
7.A
8.B
9.A
10.C hsvs whehsg

5 0

3 years ago

Which property does this example show? 2.61 + 92.7 = 92.7 + 2.61

azamat

The answer would be commutative.

Hope this helps !

Photon

6 0

3 years ago

Read 2 more answers

Consider the function.

Licemer1 [7]

Answer:

Option 4

Step-by-step explanation:

Given function is,

$f(x)=2\frac{1}{2}-3\frac{1}{3}x$

Steps to get the inverse of the given function,

Step 1,

Convert the function into equation,

$y=2\frac{1}{2}-3\frac{1}{3}x$

Step 2,

Interchange the variables 'x' and 'y',

$x=2\frac{1}{2}-3\frac{1}{3}y$

Step 3,

Solve the equation for the value of y,

$x-2\frac{1}{2}=-3\frac{1}{3}y$

$x-\frac{5}{2}=-\frac{10}{3}y$

$-x+\frac{5}{2}=\frac{10}{3}y$

$-3x+\frac{15}{2}=10y$

$-\frac{3}{10}x+\frac{15}{20}=y$

$y=-\frac{3}{10}x+\frac{3}{4}$

Step 4,

Convert the equation into inverse function,

$f^{-1}x=-\frac{3}{10}x+\frac{3}{4}$

For x-intercepts,

Substitute y = 0,

$0=-\frac{3}{10}x+\frac{3}{4}$

$\frac{3}{10}x=\frac{3}{4}$

$x=\frac{10}{4}$

$x=\frac{5}{2}$

$x=2\frac{1}{2}$

Therefore, x-intercept of the inverse function is $(2\frac{1}{2},0)$ .

Option 4 will be the answer.

8 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