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Aleonysh [2.5K]
3 years ago
15

What is the equation of the line that passes through the point (- 1, 5) and is parallel to the line y = - 3x + 5

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

<h2>          y - 5 = -3(x + 1)</h2>

Step-by-step explanation:

Parallel lines has the same slope.

The equation of a line that passing through point (x₁, y₁) with a slope of m is:

y - y₁ = m(x - x₁)

m = -3

(-1, 5)   ⇒  x₁ = -1,  y₁ = 5

Therefore the equation:

y - 5 = -3(x + 1)

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Expression n' + 5n™ + 5n 2 when n = (-3).
aksik [14]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

n' + 5n™ + 5n 2  equation

Step 2

n' + 5n™ + 5n 2  substitute n=-3

-3+(5)(-3)5(-3)2

step 3

-3+(5)(-3)5(-3)2   add and multiply

answer

27

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3 years ago
How many ways can 6 people be chosen and arranged in a straight line if there are 8 people to choose from? 720 20,160 40,320 48?
Rudik [331]
The number of permutations of 8 people taken 6 at a time is given by:
8P6=\frac{8!}{(8-6)!}=20,160&#10;
5 0
3 years ago
Read 2 more answers
For the function f(x) whose graph is shown below, what is the value of f(5)?
kipiarov [429]
F(5)=2 you look at the x-axis and find 5 which is in the right side then find the point on the line that touches and you will see it’s 2.
6 0
3 years ago
7 * 0.04 = 7 * 4 *<br> 7 times 0.04 equal 7 times 4 times ?
wolverine [178]

Answer:

Solution

1 Cancel 77 on both sides

x\times 0.04=x\times 4xx×0.04=x×4x

2 Regroup terms

0.04x=x\times 4x0.04x=x×4x

3 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

​a

​​ x

​b

​​ =x

​a+b

​​

0.04x={x}^{2}\times 40.04x=x

​2

​​ ×4

4 Regroup terms

0.04x=4{x}^{2}0.04x=4x

​2

​​

5 Move all terms to one side

0.04x-4{x}^{2}=00.04x−4x

​2

​​ =0

6) Factor out the common term x

x(0.04-4x)=0

7 Solve for x

x=0,0.01

Done!

Step-by-step explanation:

6 0
3 years ago
find the equation for the line that passes through the point (4,-3), and that is perpendicular to the line with the equation y=
Grace [21]

Answer:

hi

Step-by-step explanation:

Answer

Slope between two given points (1,3) and (2,7) is

m=  

2−1

7−3

​  

=4

Then perpendicular slope is m  

1

​  

=  

4

−1

​  

 

The line equation with this slope is given by

y=  

4

−1

​  

x+c.......(1)

Now, the above line passes through the point (−4,−3)

⇒−3=  

4

−1

​  

×(−4)+c

⇒c=−4

Therefore required line is 4y+x+16=0 (Substitute  

′

c  

′

 in (1) and simplify)

7 0
3 years ago
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