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Alja [10]
3 years ago
9

Help me please im very confused and im i wrong on my question?

Mathematics
2 answers:
g100num [7]3 years ago
5 0

It is hard to see the picture.

rjkz [21]3 years ago
3 0
U almost got it just one is incorrect in my view
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Simplify the expression <br> 9y+4-6y
Mashutka [201]

Answer:

3y+4

Step-by-step explanation:

9y-6y=3y

3 0
3 years ago
I need help with this question asap
Nataly_w [17]

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The correct answer would be the bottom equation.

Step-by-step explanation:

y=7.25x

y=7.25(3)

y=21.75

y=7.25(4)

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y=7.25(5)

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3 years ago
¡ HELP Plz !<br> Rewrite this equation in standard form.<br> y=-8x+2/7
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Eight x plus y equals two over seven

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3 years ago
Write an equation perpendicular to y=1/5 x+9 that passes through the point (-2,-2)
Mamont248 [21]

Answer:

-2= 1/5x-2 +9

-2= -0.4 +9

Step-by-step explanation:

4 0
3 years ago
Suppose each edge of the cube shown in the figure is 10 inches long. Find the sine and cosine of the angle formed by diagonals D
tatiyna

Check the picture below.

sin(EDG )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{10\sqrt{3}}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{sin(EDG )=\cfrac{\sqrt{3}}{\sqrt{3^2}}\implies sin(EDG )=\cfrac{\sqrt{3}}{3}} \\\\[-0.35em] ~\dotfill

cos(EDG )=\cfrac{\stackrel{adjacent}{10\sqrt{2}}}{\underset{hypotenuse}{10\sqrt{3}}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{cos(EDG )=\cfrac{\sqrt{6}}{\sqrt{3^2}}\implies cos(EDG )=\cfrac{\sqrt{6}}{3}}

4 0
2 years ago
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