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kotegsom [21]
3 years ago
13

Find the coordinates of the image of point R(3, -5) rotated 180 degrees about the origin.

Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0
180 degrees means that the new x=-x and the new y=-y so
x=3 and y=-5 is orignal
new is -3 and 5

so new oint is

A(-3,5)
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3.49 hectograms equals how many decigrams​
klasskru [66]

Answer:

3490 decigram

Step-by-step explanation:

1 \:hectogram = 1000 \:decigram\\\\3.49\: hectogram = x\: decigram\\\\Cross\:Muliply\\\\x = 3.49\times 1000\\\\x = 3490 \:decigram

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%20%5Cinfty%20%20%5Cfrac%7B%20%5Csqrt%5B%20%20%5Cscriptsize%5Cphi%
Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx

Replace x \to x^{\frac1\phi} = x^{\phi-1} :

I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx

Split the integral at x = 1. For the integral over [1, ∞), substitute x \to \frac1x :

\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx

The integrals involving tan⁻¹ disappear, and we're left with

I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}

8 0
2 years ago
Without using a calculator, determine the number of real zeros of the function
Lena [83]
X= 1, -3, -2 or -7 and 7
4 0
3 years ago
i haven't submitted a single assignment for this class since winter break bc i didn't realize she was assigning things on goform
rosijanka [135]

Answer and Step-by-step explanation:

<u>Trigonometric Function (These only work for right triangles):</u>

SOH-CAH-TOA

S = Sine = sin

C = Cosine = cos

T = Tangent = tan

O = Opposite (side)

H = Hypotenuse (side)

A = Adjacent (side)

SOH = sin(angle) = \frac{opposite}{hypotenuse}

CAH = cos(angle) = \frac{adjacent}{hypotenuse}

TOA = tan(angle) = \frac{opposite}{adjacent}

1. sinQ (Q is the angle) = \frac{14}{50}  = \frac{7}{25} = 0.28

Use Pythagorean Theorem to find side PQ.  14^{2} + PQ^{2}= 50^{2} \\PQ^{2}  = 2304\\PQ = 48

2. cosQ = \frac{48}{50} = \frac{24}{25} = 0.96

3. tanQ = \frac{14}{48}  = \frac{7}{24}  = 0.29

3 0
3 years ago
Joan had 198 dollars to spend on 9 books.After buying them she had 18 dollars. How much did each book cost?
Fed [463]
She spent 20 dollars on each book because if you subtract 18 from 198 you'll get 180 then you divide that by 9 n it leaves you with 20.
3 0
3 years ago
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