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Kay [80]
3 years ago
15

7x-5y=20 in slope intercept form

Mathematics
1 answer:
Leto [7]3 years ago
3 0
Slope inercept
solve for y

mnus 7x both sides

-5y=-7x+20
divide both sides by -5
y=7/5x-4
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lara31 [8.8K]

What are you stuck with

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5. Class 7887 has 2:7 math classes daily. How many math classes do they have
ANTONII [103]

Answer:

56.7

Step-by-step explanation:

2.7 x 21 = 56.7

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3 years ago
Find the lengths of the missing side . Simplify all radicals !!!<br> help mee!!!!!!
larisa86 [58]

Answer:

e = 13\sqrt{2}

f = 13\sqrt{2}

Step-by-step explanation:

The ∆ given is an isosceles ∆ with a right angle measuring 90°, and two congruent angles measuring 45° each.

Using trigonometric ratio formula, we can find the lengths of the missing side as shown below:

Finding e:

sin(\theta) = \frac{opp}{hyp}

sin(\theta) = sin(45) = \frac{\sqrt{2}}{2}

hyp = 26

opp = e = ?

Plug in the values into the formula

\frac{\sqrt{2}}{2} = \frac{e}{26}

Multiply both sides by 26

\frac{\sqrt{2}}{2}*26 = \frac{e}{26}*26

\frac{\sqrt{2}}{2}*26 = e

\frac{\sqrt{2}}{1}*13 = e

13\sqrt{2} = e

e = 13\sqrt{2}

Since side e is of the same length with side f, therefore, the length of side f = 13\sqrt{2}

3 0
3 years ago
F(x)=2x+3/4x+5<br>find f(-9)<br>​
mylen [45]

\implies {\blue {\boxed {\boxed {\purple {\sf { f(-9)= 0.48}}}}}}

\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}

f(x) =  \frac{2x + 3}{4x + 5} \\

For f(-9), put "-9" for every value of "x".

↬f( - 9) =  \frac{2( - 9) + 3}{4( - 9) + 5}\\

↬ f(-9) =  \frac{ - 18 + 3}{ - 36 + 5} \\

↬ f(-9) =  \frac{ - 15}{ - 31}\\

↬ f(-9)=  \frac{15}{31}\\

↬f(-9)=  0.48\\

\bold{ \green{ \star{ \red{Mystique35}}}}⋆

6 0
3 years ago
Read 2 more answers
Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)
Yakvenalex [24]

Answer:

\sin d = \frac{4}{7} ; \sin e = \frac{\sqrt{33} }{7}

\cos d = \frac{\sqrt{33} }{7} ; \cos e = \frac{4}{7}

\tan d = \frac{4}{\sqrt{33} } ; \tan e = \frac{\sqrt{33} }{4}

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

  • \sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \cos \alpha  = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}
  • \tan \alpha  = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }

__________________________________________________

According to the question ,

1) When α (alpha) = d

  • \sin d = \frac{4}{7}
  • \cos d = \frac{\sqrt{33} }{7}
  • \tan d = \frac{4}{\sqrt{33} }

2) When α (alpha) = e

  • \sin e = \frac{\sqrt{33} }{7}
  • \cos e = \frac{4}{7}
  • \tan e = \frac{\sqrt{33} }{4}

3 0
3 years ago
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