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3 years ago

Simplify: 4 (3y +x) please show your work

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

Answer:

12y+4x

Step-by-step explanation:

Multiply 4*3y and 4*x

Distributive property

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How must m2 be changed to keep the lines parallel?

Anna [14]

Answer:

bring the lines side by side ( example || )

Step-by-step explanation:

3 0

3 years ago

Given that triangle abc ~ def, solve for x A.6 B.3 C.2 D.4

kipiarov [429]

ABC and DEF are similar

AB/AC = DE/DF

30/24 = 5/x

30x = 24 * 5

30x = 120

x = 4

Answer

x = 4

6 0

3 years ago

Read 2 more answers

Solve; 5(x+1)=6.4−2.7x 3(y−1.5)=9.9–4.2y 0.9y−1.8=0.3(y−4)

Dmitrij [34]

—
1. Answer x=0.18

2. Answer y = 2

3. Answer y = 1

7 0

4 years ago

Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Vinvika [58]

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$

Learn more about midsegment theorem of a triangle here:

brainly.com/question/7423948

8 0

3 years ago

Help me please and ty

pav-90 [236]

Answer: 3m

And I had to write something for this to be ong enough so thats what this is:)

3 0

3 years ago