Ask question

Ask question

All categories

3 years ago

10

Plz help this is due tomarrow

2 answers:

Olin [163]3 years ago

3 0

The independent is a sentence that can stand alone a dependent isn't

lakkis [162]3 years ago

3 0

A. 2x + 14
b. I believe that x is dependent and 14 and 2 are independent

You might be interested in

The linear equation has 5(x + 2) = 5x + 2 A) no solution. B) one solution. C) two solutions. D) infinitely many solutions.

erik [133]

I would say C!!!!!!!!!

4 0

3 years ago

Read 2 more answers

SVETLANKA909090 [29]

Answer:

D

Step-by-step explanation:

can i get a brainliest? its fine i just want one ;-;

4 0

3 years ago

Read 2 more answers

Expand the function f(x)=(3x+2)

Usimov [2.4K]

Answer:

81x^4+216x^3+216x^2+96x+16

Step-by-step explanation:

6 0

2 years ago

Prove that the altitude to the base of an isosceles triangle is also the median to the base

LenKa [72]

The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

Hope It Helped! :)

4 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