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3 years ago

Find each measurement indicated. Round your answers to the nearest tenth.

Mathematics

1 answer:

Crazy boy [7]3 years ago

6 0

Answer:

Step-by-step explanation:

Find AB using the Law of Sine:

$\frac{AB}{sin(C)} = \frac{AC}{sin(B)}$

Thus:

$\frac{AB}{sin(40)} = \frac{47}{sin(103)}$

Multiply both sides by sin(40)

$\frac{AB}{sin(40)}*sin(40) = \frac{47}{sin(103)}*sin(40)$

$AB = \frac{47*Sin(40)}{sin(103)}$

AB = 31.0056916 ≈ 31.0 cm (nearest tenth)

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(6(1) = -5<br> 1) Find (4) in the sequence given by<br> (6(n) = b(n − 1) +9

Sloan [31]

Answer:

f(4) = 22

Step-by-step explanation:

I think that you type wrongly.

we have f(1) = -5

and f(n) = f(n-1) + 9

Find f(4)

Then

with n= 2 we have f(2) = f(1) + 9 = - 5+ 9= 4

with n= 3 we have f(3) = f(2) + 9 = 4 + 9= 13

and finally with n= 4 we find f(4) = f(3) + 9 = 13+ 9= 22

So f(4) = 22

Hope you understand.

8 0

3 years ago

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Can someone help me with these questions ASAP!!

Igoryamba

Hope this helps with your questions :)

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3 years ago

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Find(f−g)(x) f(x)=x2−6xandg(x)=−2x+3 <br> A x2−4x+3 <br> B x2+4x−3 <br> C x2−4x−3 <br> D x2+4x+3

3241004551 [841]

Answer:

option C is the answer......

8 0

3 years ago

Identify the expression as a numerical expression or variable expression I•8

user100 [1]

A is the answer because the number 8 is a constant. and the multiplication sign is the operator.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago