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3 years ago

8

If in the equation [ 4/t-1 = 2/w-1 ] t ≠ 1 and w ≠ 1, then t =

1 answer:

schepotkina [342]3 years ago

3 0

t=2w-1

OPTION A is the correct answer...

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A pole-vaulter uses a 15-foot-long pile. She grips the pole so that the segment below her left hand is twice the length of the s

Shalnov [3]

Let x + 1.5 be length of segment above left hand
Let 2x be length of segment below left hand

15 = x + 1.5 + 2x
15 = 3x + 1.5
15 - 1.5 = 3x
13.5 = 3x
13.5/3= 3x/3
4.5 = x
x = 4.5
This is the length above the left hand.

2x must be then 4.5*2 = 9 foot
This is the length below the left hand.

Checking
4.5 + 9 + 1.5 = 15 foot

Adding length below left hand & length of left hand to right hand:
9 + 1.5 = 10.5
The right hand is hence 10.5 foot far up the pole.

6 0

3 years ago

11. What is the reciprocal of 6/5?<br> OA. 12/20<br> OB.11/5<br> OC.1<br> OD.576

Elanso [62]

Answer: The answer is D, 5/6.

Step-by-step explanation: The reciprocal of a fraction is that fraction but the numerator and denominater swapped places.

3 0

3 years ago

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Round 832 to the nearest hundreds place.

elena-s [515]

Answer:

800

Step-by-step explanation:

Look at the tens if it's 0-4 keep the number the same if 5-9 go up one number

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3 years ago

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

8 0

3 years ago

Convert this decimal to a simplified fraction:<br><br> 0.32

alexdok [17]

The answer is 8/25
Hope it helped

3 0

3 years ago

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