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3 years ago

15

7x-5y=20 in slope intercept form

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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3 years ago

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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

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

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

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

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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 " $x$ ".

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

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

$↬ f(-9) = \frac{ - 15}{ - 31}\\$

$↬ f(-9)= \frac{15}{31}\\$

$↬f(-9)= 0.48\\$

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

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3 years ago

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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

$\sin e = \frac{\sqrt{33} }{7}$

$\cos e = \frac{4}{7}$

$\tan e = \frac{\sqrt{33} }{4}$

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3 years ago