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Rasek [7]
3 years ago
13

7y-y=30 show work for it!

Mathematics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

y=5

Step-by-step explanation:

7y-y=6y

30/6 = 6y/6

5 = y

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Graph this function y-2=-(x+5)
I am Lyosha [343]
Going to put your function in slope-intercept form for the sake of easier input into Desmos.

y - 2 = - (x + 5)
y - 2 = - x - 5
y = - x - 5 + 2
y = - x - 3


4 0
4 years ago
Suppose △LMN≅△PQR .
Tanya [424]
First four answers are correct as congruent parts of congruent triangles are congruent but L Is not the opposite to M so they wouldn’t be congruent nor does NL or PQ have the same connection
6 0
3 years ago
Matthew wants to save up to buy a new PS5 whenever it comes out. He
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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1. Draw a histogram from all the data. Starting at the bottom row, for each set of 10 flips, place an “x” in the column correspo
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Answer:

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Step-by-step explanation:

4 0
4 years ago
Suppose that the terminal side of angle alphaα lies in Quadrant I and the terminal side of angle betaβ lies in Quadrant IV. If s
melamori03 [73]

Solution :

It is given that :

$\alpha$ lies in the first quadrant.

And $\beta$ lies in the fourth quadrant.

Since, $\sin \alpha = \frac{5}{13}$     and $\cos \beta = \frac{6}{\sqrt{85}}$    (given)

$\sin \alpha = \frac{5}{13}$  

$\cos \alpha = \sqrt{1-\sin^2 \alpha}$

   $\cos \alpha = \frac{12}{13}$

Similarly  $\cos \beta = \frac{6}{\sqrt{85}}$

$\sin \beta = \sqrt{1-\cos^2 \beta}$

$\sin \beta = \sqrt{1-\frac{36}{85}}$

     $-\frac{7}{\sqrt{85}}$      (IVth quadrant)

Therefore,

$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

                 $=\frac{12}{13}\times \frac{6}{\sqrt{85}}-\frac{5}{13}\times \frac{-7}{\sqrt{85}}$

                $= \frac{107}{13 \sqrt{85}}$

6 0
3 years ago
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