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2 years ago

50 POINTS HELP ASAP PLZ What is the equation of this line?

Mathematics

1 answer:

RSB [31]2 years ago

6 0

Answer:

i think its the third one

Step-by-step explanation: im srry if im wrong doe

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The yearbook team is designing the pages for school clubs so

yarga [219]

Answer:

creo que sería 54 por actitud de :72-1 1?? por eso sería esta cosa esxtactamente espero que te allá ayudado

6 0

2 years ago

Can you please help please please help me with this question is so hard and i will give you a Brainiest please

mrs_skeptik [129]

Question 1 :

x ≤ 4

Question 2:

i think its 40-5

7 0

2 years ago

Asap please and thank you

fgiga [73]

Answer:

<h2> x = 16 degrees</h2>

Step-by-step explanation:

1 and 3 are <u>vertical angles</u> so m∠1 = m∠3

(4x - 2)° = 62°

4x - 2 = 62

4x = 64

x = 16

5 0

3 years ago

Hey mate answer te question please

andrezito [222]

Answer:

7 $\frac{37}{55}$

Step-by-step explanation:

In order to solve an equation like this, you'll need to find a common denominator for your Fraction.

7 $\frac{2}{5}$ + $\frac{3}{11}$

7 $\frac{22}{55}$ + $\frac{15}{55}$

7 $\frac{37}{55}$

4 0

3 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

2 years ago