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3 years ago

Suppose triangle A B C is similar to triangle P Q R. Also, angle A measures 45° and angle B measures 60°.

1 answer:

4 0

Answer: It looks like you are missing the question here. Most of the time in this type of questions, we are looking for the 3rd angle.

Let's assume that you need the 3rd angle.

There are 180 degrees in a triangle. We know 2 of the angles, so by subtracting them from 180 we can find the 3rd angle.

180 - 45 - 60 = 75

Angle C is 75 degrees. In the similar triangle Angle R is also 75 degrees because the corresponding angles are congruent.

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I really need help. Find a. Round to the nearest tenth.

valina [46]

Answer:

a ≈ 1.8

Step-by-step explanation:

a / sin (180 - 105 - 15)° = 2 /sin 105°

a = (2 /sin 105°) x sin 60°

a = (2 / 0.97) x 0.87

a = 1.79 (≈ 1.8)

6 0

2 years ago

$1876 per month for 12 months<br> A) $22,512<br> B) $225.12<br> c) $1888<br> D) None of these

insens350 [35]

Answer:

The answer is A

Step-by-step explanation:

We need to multiply 1876 by 12. 1876 * 12 = 22, 512. That's our answer

5 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

2 years ago

I'm really lost on this problem

Alenkinab [10]

Since you need a little bit of trigonometry to solve this problem,
I'm pretty sure you've had a little bit of trigonometry in class before
you were assigned to solve this problem.

-- The shaded triangle comes from taking the equilateral triangle and
either folding it in half or cutting it in half.

-- The base that the new triangle is standing on is 1/2 the base of the
equilateral triangle, so it's 7 cm long.

-- The acute angle at the right end of the base is the same as it was in
the equilateral triangle . . . 60 degrees.

-- The tangent of 60 degrees is (opposite side)/(adjacent side) = x / 7 cm

tan(60) = x/7

-- Multiply each side of this little equation by 7 : 7 tan(60) = x

If you don't have the tangent of 60 degrees in your pocket, you can
find it with your calculator. Multiply it by 7, and Shazam, you know 'x' .

5 0

2 years ago

Celina measured the height of a window frame as 5.55 feet, but the actual height was 6 feet. What is the percentage of error in

REY [17]

Is this Dolli Buchannan Its Alexis!!!!

8 0

2 years ago