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3 years ago

PLEASE HURRY, MARKING BRAINLIEST.

Mathematics

2 answers:

Hitman42 [59]3 years ago

8 0

X = 16
You set the equations equal and then solvep

Tomtit [17]3 years ago

5 0

Answer:

x=16

Step-by-step explanation:

if the lines were parallel then that means that (7x+2) = (8x-14) because they are corresponding angles. just isolate the varible so subtract 7x from both sides and add 14 to both sides. and they you get x=16

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Can yall help me on question 23?!

Allushta [10]

Answer:

c. 29 feet

Step-by-step explanation:

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3 years ago

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Nataliya [291]

Answer:

12 - B

13 - No

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

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3 years ago

What are the answers to A and C?

damaskus [11]

8x+1=-x-1 4-5x=1+6x
8x+x=-1-1 -5x-6x=1-4
9×=-2 -11×=-3
×=-2/9 ×=3/11

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3 years ago

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lbvjy [14]

Answer:

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Step-by-step explanation:

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3 years ago