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Arte-miy333 [17]
3 years ago
8

Write an equation for a line that passes thruogh (3,-1) and has a slope of 2

Mathematics
1 answer:
harkovskaia [24]3 years ago
3 0

Answer:

y=2x−7

Step-by-step explanation:

so

formula is ( y 1 − y = m ( x 1 − x))

m is the slope

-1-y=2(3-x)

add 1 to both sides

−y−1+1=−2x+6+1

−y=−2x+7

divide both sides by -1

y=2x−7

Hope this helps

Brainliest please

Ask if u have more questions.

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Eight less than four times a number is less than 56. What are the possible<br> Values of that number
Ivan

Answer:

Step-by-step explanation:

If you would like to write the following in the math expression, you can do this using these steps:

a number ... n

eight less than four times a number ... 4 * n - 8

is less than 56 ... < 56

4 * n - 8 < 56

4 * n < 56 + 8

4 * n < 64     /4

n < 64 / 4

n < 16

The possible values of that number are less than 16.

7 0
4 years ago
Calculate the shaded region
Nikitich [7]

Answer:

196 cm²-153.9375 cm² = 42.0625 cm ²

Step-by-step explanation:

First, solve for the whole 14^2 = 196 cm²

Now, we know it's lenths are the same so it's a square.

Lastly we need to deduct the quarter circle in the square. Use Formula A=πr² = A≈615.75 cm² /4 = 153.9375 cm²

5 0
3 years ago
-5(n + 3) = 6(n +4) + 5
laiz [17]

Answer:

n = -4

Step-by-step explanation:

−5(n+3)=6(n+4)+5

Step 1: Simplify both sides of the equation.

−5(n+3)=6(n+4)+5

(−5)(n)+(−5)(3)=(6)(n)+(6)(4)+5(Distribute)

−5n+−15=6n+24+5

−5n−15=(6n)+(24+5)(Combine Like Terms)

−5n−15=6n+29

−5n−15=6n+29

Step 2: Subtract 6n from both sides.

−5n−15−6n=6n+29−6n

−11n−15=29

Step 3: Add 15 to both sides.

−11n−15+15=29+15

−11n=44

Step 4: Divide both sides by -11.

−11n

−11

=

44

−11

n=−4

6 0
4 years ago
Read 2 more answers
3 5 1 8
Neporo4naja [7]
The answer is 7173
8531-1358
3 0
3 years ago
Sin A tan A/ 1 - cos A = 1+ sec A​
aalyn [17]

\text{L.H.S}\\\\=\dfrac{\sin A \tan A}{1- \cos A}\\\\\\=\dfrac{\sin A \cdot \tfrac{\sin A}{\cos A} }{ 1- \cos A}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~;\left[\tan A = \dfrac{\sin A }{\cos A}\right]\\\\\\=\dfrac{\tfrac{\sin^2 A}{ \cos A}}{1- \cos A}\\\\\\=\dfrac{\sin^2 A}{\cos A(1-\cos A)}\\\\\\=\dfrac{1-\cos^2 A}{\cos A( 1- \cos A)}~~~~~~~~~~~~~~~~~~~~~~~~;[\sin^2 A + \cos^2 A = 1]\\\\\\

=\dfrac{(1+ \cos A)(1 - \cos A)}{\cos A(1 - \cos A)}\\\\\\=\dfrac{1+ \cos A}{ \cos A}\\\\\\=\dfrac{1}{\cos A} + \dfrac{\cos A}{ \cos A}\\\\\\=\sec A + 1 \\\\\\=1+ \sec A\\\\\\=\text{R.H.S}\\\\\text{Proved.}

8 0
2 years ago
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