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3 years ago

Write an equation parallel to the line determined by the points (15, -6) and (-3, 13), through: (4, 2)

Mathematics

2 answers:

gulaghasi [49]3 years ago

7 0

Answer:

The answer is

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

To find the equation of the parallel line we must first find the slope of the original line

That's

Slope of the through points

(15, -6) and (-3, 13) is

Since the lines are parallel their slope are also the same

So slope of parallel line = - 19/18

Equation of the line using point (4,2) and slope -19/18 is

We have the final answer as

Hope this helps you

galben [10]3 years ago

6 0

Answer:y=-1.583x-8.332

Step-by-step explanation:

First find slope from two points (-6-13)/(15+3)=-1.583

Now line is parallel so the slope would be same for the other line passing through (4,2) now as the general equation of line is

y=mx+c

2=-1.583(4)+c

Solving for c equals to -8.332

So final equation is

y=1.583x-8.332

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Simplify this expression; 1/2s - t + 3s - 1/4 t

Svetach [21]

Answer:

14s-5t/4

Step-by-step explanation:

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3 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

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3 years ago

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kirill [66]

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3 years ago

Solve this equation for A:

12345 [234]

a₁₁ = -11

a₁₂ = -16

a₁₃ = 6

a₁₄ = -10

Step-by-step explanation:

The matrix A will therefore appear as;

[ -11 -16 6 -10 ]

Adding matrices of the same size involves adding the individual elements;

A = [ 0 1 -2 3 ] – [ 11 17 -8 13 ]

Substance the individual element in the two matrices that share the same coordinates;

0 – 11 = -11

1 – 17 = -16

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3 – 13 = -10

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