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const2013 [10]
3 years ago
7

If f(x) =4x-2 and g(x) =2x+8, what is h (x) when h (x) =f (x) +g(x)

Mathematics
2 answers:
melisa1 [442]3 years ago
8 0

Answer:

h(x)=4x_2+2x+8

h(x)=6x+6

uysha [10]3 years ago
8 0

<u>Answer:</u>

<h2>h(x) = 6x + 6</h2>

<u>Steps:</u>

given:

f(x) = 4x-2

g(x) = 2x+8

h(x) = f(x) + g(x)

h(x) = 4x-2 + 2x+8

h(x) = 4x + 2x + 8 - 2

h(x) = 6x + 6

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Write your answer as a while number
rusak2 [61]

Answer:

10 is your answer

Step-by-step explanation:

2×15/3

= 10

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2 years ago
Need help with this question
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3 years ago
4i+4-2i+10<br> whats the answer to this equation 0_0
N76 [4]

Answer:

2i + 14

Step-by-step explanation:

4i + 4 - 2i + 10

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3 0
3 years ago
Read 2 more answers
Write a formula that will yield the nth term of the sequence: 3,-2,-7,-12
RUDIKE [14]

Answer:

-5n + 8

Step-by-step explanation:

<u>1. What is the difference?</u>

  The sequence goes down by 5. This means that the formula will have -5n in it.

<u>2. Work out the term before the first term.</u>

    The first term is 3, and we know that the sequence goes up by 5. So, to get the term before 3 we would add 5.

3 + 5 = 8   Remember, this is positive 8. (+8)

<u>3. Put that term at the end of the equation. </u>

   We have -5n already and we just worked out the term before the first one which is positive 8 - so put that at the end of the equation.

-5n + 8  This is our answer!

Just to prove it works:

<em>Substitute:   n = term</em>

Lets see if we can get the 3rd term which is -7. (n = 3)

-5(3) + 8

-15 + 8 = -7

See, it works!

8 0
3 years ago
Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet
zmey [24]

Answer:

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

hypotenuse^2=perpendicular^2+base^2

hypotenuse^2=(5)^2+(12)^2

hypotenuse^2=25+144

hypotenuse^2=169

Taking square root on both sides.

hypotenuse=13

In a right angled triangle

\sin \theta = \dfrac{opposite}{hypotenuse}

\sin \theta = \dfrac{5}{13}

\sec \theta = \dfrac{hypotenuse}{adjacent}

\sec \theta = \dfrac{13}{12}

In second quadrant only sine and cosecant are positive.

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

6 0
3 years ago
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