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3 years ago

What is m∠2? What is m∠1?

Mathematics

2 answers:

Greeley [361]3 years ago

6 0

Answer:

<2 = 75

<1 = 105

Step-by-step explanation:

<2 and 75 are alternate interior angles and since the lines are parallel, they are equal

<2 = 75

<2 + <1 = 180 since the angles form a line

75+ <1 = 180

Subtract 75 from each side

<1 = 180-75

<1 = 105

MA_775_DIABLO [31]3 years ago

6 0

Answer:

105

Step-by-step explanation:

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Cathy uses 3/4 teaspoon of vanilla in a batch of cookies. How many teaspoons are needed for 8 batches of cookies

34kurt

6 teaspoons
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3 years ago

What is y=2x+4 and 3x+y=9 substitution method

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3 years ago

Solve this thing pls <img src="https://tex.z-dn.net/?f=3%5Cleft%28q-7%5Cright%29%3D27" id="TexFormula1" title="3\left(q-7\ri

evablogger [386]

Answer:

Solution of equation ( q ) = 16

Step-by-step explanation:

In this question we have given an equation that is 3 ( q - 7 ) = 27 and we have asked to solve this equation that means to find the value of  q .

Solution : -

$\quad \: \longmapsto \: 3(q - 7) = 27$

Step 1 : Solving parenthesis :

$\quad \: \longmapsto \:3q - 21 = 27$

Step 2 : Adding 21 on both sides :

$\quad \: \longmapsto \:3q - \cancel{ 21} + \cancel{21} = 27 + 21$

On further calculations we get :

$\quad \: \longmapsto \:3q = 48$

Step 3 : Dividing by 3 from both sides :

$\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}} = \cancel {\frac{48}{3} }$

On further calculations we get :

$\quad \: \longmapsto \: \pink{\boxed{\frak{q = 16}}}$

Therefore, solution of equation ( q ) is 16 .

Verifying : -

Now we are very our answer by substituting value of q in the given equation . So ,

3 ( q - 7 ) = 27

3 ( 16 - 7 ) = 27

3 ( 9 ) = 27

27 = 27

L.H.S = R.H.S

Hence, Verified.

Therefore, our solution is correct .

<h2>#Keep Learning</h2>

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2 years ago

Read 2 more answers

Help!! 20points! thank you need by tomorrow.

Yakvenalex [24]

Answer:

1. f(x,y) = (x, -y) 2. f(x,y) = (-x,y).

Step-by-step explanation:

I hope that helps

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3 years ago

Is sin theta=5/6, what are the values of cos theta and tan theta?

mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

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3 years ago