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3 years ago

Look at the following function X -12 -8

Mathematics

1 answer:

Julli [10]3 years ago

6 0

X-12 -8
calculate the difference
X-20

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MARKING BRAINIEST ANSWER!!!

cluponka [151]

Answer:

y = 3

Step-by-step explanation:

Hello,

$f(x)=(\dfrac{1}{2})^x+3$

when x tends to $+\infty$

$(1/2)^x$ tends to 0 as 1/2 < 1

so the equation of the asymptote for this function is y = 3

3 0

3 years ago

Read 2 more answers

Help and i'll give you brainliest

8090 [49]

Answer:

139°

Step-by-step explanation:

They would both be 139

6 0

3 years ago

Which sequence matches the recursive formula? an=2an-1+5, where a1=5. A) 5,10,15,20 B) 5,15,35,75. C)5,15,25,35. D) 5,20,35,40

Dominik [7]

The recursive formula

$a_n = 2a_{n-1}+5$

means that, to compute the next value, you have to double the previous one and then add 5.

So, we start with 5. To get the second term, we double it and add five, so we have

$a_2 = 2\cdot 5+5 = 10+5 = 15$

Next, we will double 15 and add 5:

$a_3 = 2\cdot 15 + 5 = 30+5 = 35$

So, the answer is B, because it's the only option starting with 5,15,35.

Just to check, let's compute the last step:

$a_4 = 2\cdot 35 + 5 = 70+5 = 75$

Which confirms option B

6 0

3 years ago

Find the exact solution of 3x^2+7=28

Softa [21]

$\text{Solve:}\\\\3x^2+7=28\\\\\text{Subtract 7 from both sides}\\\\3x^2=21\\\\\text{Divide both sides by 3}\\\\x^2=7\\\\\text{Square root both sides}\\\\\sqrt{x^2}=\sqrt7\\\\x=\pm\sqrt7\\\\\boxed{x=\sqrt7\,\,or\,\,x=-\sqrt7}$

4 0

3 years ago

What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

crimeas [40]

Answer:

$\large\boxed{x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}}$

Step-by-step explanation:

$f(x)=2x^2+8x-3\\\\\text{You can use the quadratic formula:}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ a=2,\ b=8,\ c=-3\\\\\text{Substitute:}\\\\b^2-4ac=8^2-4(2)(-3)=64+24=88\\\\\sqrt{b^2-4ac}=\sqrt{88}=\sqrt{4\cdot22}=\sqrt4\cdot\sqrt{22}=2\sqrt{22}\\\\x=\dfrac{-8\pm2\sqrt{22}}{2(2)}=\dfrac{-8}{4}\pm\dfrac{2\sqrt{22}}{4}=-2\pm\dfrac{\sqrt{22}}{2}=-2\pm\dfrac{\sqrt{22}}{\sqrt4}\\\\x=-2\pm\sqrt{\dfrac{22}{4}}=-2\pm\sqrt{\dfrac{11}{2}}$

5 0

3 years ago

Read 2 more answers