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makvit [3.9K]
2 years ago
10

Please help with both! :)

Mathematics
1 answer:
Vaselesa [24]2 years ago
4 0

Answer:

Step-by-step explanation:

for table the equation is

A)   y = 2x + 5

for graph equation is

y = 1/2x -1

You might be interested in
Find the equation of the line that is perpendicular to y = - 1/3 x + 2 and passes though the point (-5,2)
elena55 [62]

Answer:

y - 2 = 3(x + 5)

Step-by-step explanation:

The given equation has slope -1/3.  Any line perpendicular to the given line has a slope which is the negative reciprocal of -1/3, which comes out to +3.

Use the point-slope formula y - k = m(x - h):

y - 2 = 3(x + 5)

6 0
3 years ago
Given the following functions, find and simplify (f • g)(x) f(x)=—x+3 g(x)=-x-1
Hitman42 [59]

Answer:

(f\circ g)(x)= x + 4

Step-by-step explanation:

<u>Composite Function</u>

Given f(x) and g(x) as real functions, the composite function (f\circ g)(x) is defined as:

(f\circ g)(x)=f(g(x))

It can be found by substituting g into f.

We are given the functions:

f(x) = -x + 3

g(x) = -x -1

Find

(f\circ g)(x)=f(g(x))= - (-x -1) + 3

Removing parentheses:

(f\circ g)(x)= x + 1 + 3

Simplifying:

\mathbf{(f\circ g)(x)= x + 4}

5 0
3 years ago
3 2/5 + 9 1/10= Simplest form
Talja [164]

Change each mixed fraction to improper

3 2/5 = 15/5 + 2/5 = 17/5

9 1/10 = 90/10 + 1/10 = 91/10

Find a common denominator. Multiply 2 to both numerator and denominator of 17/5

17/5(2/2) = 34/10

34/10 + 91/10 = 125/10

125/10

Simplify. Divide

125/10 = 12.5

12.5 is your answer

hope this helps

3 0
2 years ago
Read 2 more answers
Please please please help me thanks
goldfiish [28.3K]

Answer:

1. 57       2. 81

Step-by-step explanation:

71x2=142   256-142=114/2= 57

5103/63= 81

5 0
2 years ago
Between which two square roots of integers could you find π
LUCKY_DIMON [66]
Hello.

Since 3 < pi < 4, 
√9 < pi √16 

In fact, since pi^2 = 9.86, 

<span>√9 < pi < √10. 

Which means the you</span><span> can find pi between square roots √9 and √10.
</span>
Have a nice day
5 0
2 years ago
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