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ella [17]
3 years ago
12

7. Which answer is equal to (f – g)(5) or f(5) – g(5), given f(x) = x + 2 and g(x) = x² + 5?

Mathematics
1 answer:
erastova [34]3 years ago
6 0
Hello:
<span> (f – g)(5) = f(5) – g(5) = (5+2)-((5)²+5) = 7-30
</span> (f – g)(5) = -23

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Point V is on line segment \overline{UW}
____ [38]

Answer:

UV=25 units

Step-by-step explanation:

we know that

UW=UV+VW -----> by addition segment postulate

substitute the given values

4x+10=5x+5

solve for x

5x-4x=10-5

x=5

<em>Find the value of UV</em>

UV=5x

substitute the value of x

UV=5(5)=25 units

5 0
3 years ago
Help Asap Please!! mark as Brainliest
sdas [7]

first is distributive property

next is the subtraction property of equality

then you simplify

finally you use the division property of equality

hope this helps  

4 0
3 years ago
1) Is this graph a linear function? Explain.....2) Is this graph a linear function? Explain....
Sav [38]

9514 1404 393

Answer:

  1. no
  2. no

Step-by-step explanation:

If the graph is not a straight line, it is not a linear function. Neither graph 1 nor graph 2 is a straight line, so neither is a linear function.

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Graph 2 is made of parts of two straight lines. It is not one straight line, so not a linear function. It is called "piecewise linear," which is different.

6 0
3 years ago
Solve for x<br> pls show work<br> 1/2x + 4 = 1/8x + 88
SVETLANKA909090 [29]

Answer:

224

Step-by-step explanation:

1/2x+4=1/8x+88

1/2x-1/8x+4=88

1/2x-1/8x=88-4

4/8x-1/8x=84

3/8x=84

x=84/(3/8)

x=(84/1)(8/3)

x=672/3

x=224

Please mark me as Brainliest if you're satisfied with the answer.

3 0
3 years ago
What is the correct first step to solve this system of equations by elimination?
storchak [24]

\bold{\huge{\pink{\underline{ Solution }}}}

\bold{\underline{ Given }}

  • <u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

\bold{\underline{ To \: Find }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>

\bold{\underline{ Let's \: Begin }}

\sf{ 2x - 3y = -6 ...eq(1)}

\sf{ x +  3y = 12 ...eq(2)}

<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>

\sf{ 2( x + 3y = 12 )}

\sf{ 2x + 6y = 24 }

<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}

\sf{ 2x + 6y - 2x + 3y = 24 + 6 }

\sf{   9y = 30 }

\sf{   y = 30/9}

\sf{\red{ y = 10/3}}

<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

\sf{ 2x - 3(10/3) = -6 }

\sf{ 2x - 10 = -6 }

\sf{ 2x  = -6 + 10}

\sf{ x  = 4/2}

\sf{\red{ x  = 2}}

Hence, The value of x and y is 2 and 10/3

6 0
3 years ago
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