2 years ago

What is the slope of the line x=4 ?

Mathematics

2 answers:

hodyreva [135]2 years ago

7 0

There’s no slope because it’s a vertical line.

ratelena [41]2 years ago

3 0

Answer:

Step-by-step explanation:

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Analyze the following pattern 1,2,5,10,7

vodka [1.7K]

Answer:

an + 1 = an + 2n -1 ,a 1=1

a8= 50

Step-by-step explanation:

an+1=an+2n-1,a1=1

a6=a5+2x5-1

a6=26

a7=26+11

a7=37

a8=37+13

a8=50

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3 years ago

Pl help it’s for a grade

Alla [95]

Answer:

Step-by-step explanation:

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3 years ago

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If f(x)=x+7 and g(x)=1/x-13, what is the domain of (f o g)(x)

astra-53 [7]

(f o g)(x) = 1/(x - 13) +7
domain of the function is all real x except x = 13

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3 years ago

I dont know ? 6y2 + 6(4y2 − 8) =

RideAnS [48]

Answer:

6y2 + 6(4y2 - 8)

6y2 + 24y2 - 48

30y2 - 48

Step-by-step explanation:

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3 years ago

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Using the U- Substitution u=sqrt(2x), integral form 2-8 dx/ sqrt(2x) + 1 is equivalent to ...

AleksAgata [21]

We will use u-substitute: $u= \sqrt{2x} , \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u}$ Then for substitution:dx=u du. and integral becomes: $\int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx$ =u-ln(u+1)= $\sqrt{2x}-ln( \sqrt{2x}+1)$ . Now we will change the values of limits: $\sqrt{16}-ln( \sqrt{16}+1)-( \sqrt{4}-ln( \sqrt{4}+1))$ =4-ln(5)-2+ln(3)=2+ln(0,6)=2-0.51=1.49

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3 years ago