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3 years ago

Find four consecutive even integers with a sum of -52

Mathematics

2 answers:

Komok [63]3 years ago

7 0

Answer:

-16,-14,-12,-10

Step-by-step explanation:

Let the integers be x, x+2,x+4,x+6

ATQ,

x+x+2+x+4+x+6=-52

4x+12=-52

4x=-52-12

x=-64/4

x=-16

Amiraneli [1.4K]3 years ago

3 0

Step-by-step explanation: In this problem we have four consecutive even integers that have a sum of -52. Consecutive even integers can be represented as followed.

X

X + 2

X + 4

X + 6

Since the sum of these is -52, we can set up an equation to represent this.

X + X + 2 + X + 4 + X + 6 = -52

4x + 12 = -52

4x = -64

X = -16

X + 2 = -14

X + 4 = -12

X + 6 = -10

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What is the slope of a line containing the points (12, 6) and (3, -5)?

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3 years ago

What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

maks197457 [2]

Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

$y=\dfrac{4}{2}$

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

3 0

3 years ago