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2 years ago

Plz help me with these math questions

Mathematics

1 answer:

Akimi4 [234]2 years ago

8 0

For b on the first paper you will start with the distribution property and multiply both x and 2 by 3 because they are in the ()
now you have 3x+ 6=x-18
from here you can subtract x from both sides
2x+6=-18
next subtract 6 from both sides
2x=-24
divide both sides by 2...
x=-12
hope this helped

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If LM = 4x, MN = x + 1, and LN = 6, what is MN?

Ilia_Sergeevich [38]

The value of the MN is 2 according to the assumption on the number line.

According to the statement

We have to find that the value of the MN with the help of the value of the x.

So, For this purpose, we know that the

A number line is a horizontal line that has equally spread number increments.

Now,

Assuming that L, M, and N are points on a number line with L < M < N,

According to the assumption, the equation become

4x + x+1 = 6

5x +1 =6

5x =5

x = 1.

Now, the value of the MN become:

MN = x + 1

MN = 1 + 1

MN = 2.

So, The value of the MN is 2 according to the assumption on the number line.

Learn more about number line here

brainly.com/question/4727909

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2 years ago

For what value of a should you solve the system of elimination?

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$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
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$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

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$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

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$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
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$60\left(-a+2\right)$

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$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

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GaryK [48]

Answer:

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Step-by-step explanation:

Through point O draw a ray on left side of O which is || to AB & CD and take any point P on it.

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∠ABO + ∠BOP = 180° (by interior angle Postulate)

118° + ∠BOP = 180°

∠BOP = 180° - 118°

∠BOP = 62°.... (1)

Since, ∠BOP + ∠POD = ∠BOD

Therefore, 62° + ∠POD = 152°

∠POD = 152° - 62°

∠POD = 90°.....(2)

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90° + ∠ODC = 180°

∠ODC = 180° - 90°

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