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GarryVolchara [31]

3 years ago

10

4(2+3c)=56how woud i solve this Multi-step equation sow with all

2 answers:

GaryK [48]3 years ago

7 0

4(2+3c)=56
8+12c=56
-8 -8
12c=48
/12 /12
c=4

grigory [225]3 years ago

3 0

$4(2+3c)=56\ \ \ \ |omiting\ the\ bracket\\\\
8+12c=56\ \ \ \ |subtract\ 8\\\\12c=48\ \ \ |divide\ by\ 12\\\\
c=4\\\\Solution\ is\ c=4.$ Answer here

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Consider the graph of the linear function h(x) = –x + 5. Which could you change to move the graph down 3 units?

Yakvenalex [24]

I don't see a b or an m in this function, but generally, a linear function is of the form ax+b. The b in this equation (not sure if it is the same as your b), controls the shift of the function. Lower b by 3, the graph shifts down by 3.

So unless there is a minus sign before b, it has to be lowered by 3. (If there is a minus sign, it has to be increased by 3).

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2 years ago

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Orlov [11]

The answer would be:

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3 years ago

A rectangular sheet of paper is 25/2cm long and 32/3 cm wide. Find its perimeter. *

Zina [86]

Answer:

46.3cm

Step-by-step explanation:

perimeter is length around shape
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2 years ago

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1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Sveta_85 [38]

Answer:

1. $-5x+3y+44=0$

2. $2x+y-2=0$

3. $2x+y-4=0$

Step-by-step explanation:

Standard form of a line is $Ax+By+C=0$ .

If a line passing through two points then the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope, i.e., $m=\dfrac{y_2-y_1}{x_2-x_1}$ .

1.

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

$y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)$

$y+3=\dfrac{-5}{-3}(x-7)$

$y+3=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$-5x+3y+9+35=0$

$-5x+3y+44=0$

Therefore, the required equation is $-5x+3y+44=0$ .

2.

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to $2 x + y =-5$ .

Slope of the line : $m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2$

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

$y-2=-2(x-0)$

$y-2=-2x$

$2x+y-2=0$

Therefore, the required equation is $2x+y-2=0$ .

3.

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to $2x + y =-5$ .

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

$y-0=-2(x-2)$

$y=-2x+4$

$2x+y-4=0$

Therefore, the required equation is $2x+y-4=0$ .

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3 years ago

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laila [671]

Answer:

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Step-by-step explanation:

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3 years ago