Ask question

Ask question

All categories

Ilia_Sergeevich [38]

2 years ago

11

2a - 10 = 5(4+.2a) Linear equations

1 answer:

xxMikexx [17]2 years ago

8 0

Step-by-step explanation:

2a-10=5(4+.2a)

Use the distributive property to multiply 5 by 4+0.2a..

2a-10=20+a

Subtract a from both sides.

2a-10-a=20

Combine 2a and -a to get a.

a-10=20

Add 10 to both sides.

a=20+10

Add 20 and 10 to get 30.

a=30

You might be interested in

The street fair charges an admission price of $5 per person. In the first hour, one gate admitted 10 people. In the second hour,

nasty-shy [4]

Answer:

Step-by-step explanation:

a

3 0

3 years ago

Read 2 more answers

Write an expression from the description. Then

vodka [1.7K]

(-36 ÷ -4) + 9

The expression equals 18.

4 0

3 years ago

You buy two types of fish at the local market. You need 1.5 pounds of tilapia and

Mila [183]

Answer:

$9.35

Step-by-step explanation:

Tilapia: 1.5 * $3.88 = $5.82

Cod: 1 * 3.53 = $3.53

$3.53 + $5.82 = $9.35

6 0

3 years ago

Read 2 more answers

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

<u>Find the distance AB</u>

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

<u>Find the distance BC</u>

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

<u>Find the distance AC</u>

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

<u>Find the distance DC</u>

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

<u>Find the perimeter of the triangle</u>

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

<u>Find the area of the triangle</u>

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

4 0

3 years ago

Read 2 more answers

Help my paper is almost ripped and I don't now 12 and 14

choli [55]

The word missing in 12 is fraction and field
the missing word in 14 is dividing

5 0

2 years ago