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3 years ago

Help!!! IXL! plzzzzzz.Quick

Mathematics

2 answers:

Blababa [14]3 years ago

6 0

x = 7

8x - 8y = 16.

Plug 7 from the first equation into x for the second equation

8(7) - 8y = 16.

Simplify the left side.

56 - 8y = 16.

Subtract 56 from each side.

-8y = -40.

Divide each side by -8

y = 5.

We know x = 7 and y = 5, therefore,

Your answer should be (7,5)

valentinak56 [21]3 years ago

4 0

The answer is (7, 5)

X=7

8x-8y=16

8*7-8y=16

56-8y=16

-8y=-40

y=5

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A line segment has endpointsA(-5, 1) and b(-3, 3) what is the midpoint of the segment

Galina-37 [17]

The mid point of the segment would be (-4,2)

3 0

3 years ago

Read 2 more answers

Through(-2,-3); parallel to 3x+2y=5

Vilka [71]

Step-by-step explanation:

Hey there!

Firstly find slope of the given equation.

Given eqaution is: 3x + 2y = 5.......(i)

Now;

$slope(m1) = \frac{ - coeff. \: of \: x}{coeff. \: of \: y}$

$or \: m1 = \frac{ - 3}{2}$

Therefore, slope (m1) = -3/2.

As per the condition of parallel lines,

Slope of the 1st eqaution (m1) = Slope of the 2nd eqaution (m2) = -3/2.

The point is; (-2,-3). From the above solution we know that the slope is (-3/2). So, the eqaution of a line which passes through the point (-2,-3) is;

(y-y1) = m2 (x-x1)

~ Keep all values.

$(y + 3) = \frac{ - 3}{2} (x + 2)$

~ Simplify it.

$2(y + 3) = - 3x - 6$

$2y + 6 = - 3x - 6$

$3x + 2y + 12 = 0$

Therefore, the eqaution of the line which passes through the point (-2,-3) and parallel to 3x + 2y= 5 is 3x + 2y +12 =0.

Hope it helps...

8 0

2 years ago

Read 2 more answers

I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t

belka [17]

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

$\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}$

Where

$\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}$

Thus,

$\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}$

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

$\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}$

Thus, from the theorem of similar triangles,

$\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}$

solving for y, we have

$\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}$

thus, YC = 10.