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7nadin3 [17]
3 years ago
8

A circle has a radius of 5 ft, and an arc of length 7 ft is made by the intersection of the circle with a central angle. Which e

quation gives the measure of the central angle, q?
Mathematics
2 answers:
3241004551 [841]3 years ago
8 0

arc length, s=(radius, r)(angle in radians, q)

q=tan^-1(s/r)

SO that saying the answer is 1.4 radians and the equation is probs q=7/5=1.4

WINSTONCH [101]3 years ago
8 0

Answer:

I think it's theta = 7/5

Step-by-step explanation:

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What is the measure of Angle B
patriot [66]

Answer:

119

Step-by-step explanation:

Angle C= 62

So you add 62 + 57 which gives you 119

All angles must add up to 180 so you subtract 119 from 180 which gives you 61

Since angle A is 61 and a straight line angle is 180 you subtract 61 from 180 to give you angle B

180 - 61 = 119

therefore your answer is 119

4 0
3 years ago
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Sewing patterns recommended buying extra fabric in case of mistakes. Darlene bought 7 yards of fabric, plus 20% extra fabric. Ch
Debora [2.8K]

Answer:

Darlene bought more fabric

Darlene bought 3.4 yards more

Step-by-step explanation:

Darlene = 7 yards + 20% extra

= 7 yards + (20% of 7 yards)

= 7 + (20% × 7)

= 7 + (0.2 × 7)

= 7 + 1.4

= 8.4 yards

Chris = 4 yards + 25% of 4 yards

= 4 yards + (25% of 4 yards)

= 4 + (25% × 4)

= 4 + (0.25 × 4)

= 4 + 1

= 5 yards

Darlene bought more fabric

How much more?

Darlene - Chris

= 8.4 yards - 5 yards

= 3.4 yards

Darlene bought 3.4 yards more

6 0
3 years ago
Susan deposited $275 in a savings account that earns simple interest at a rate of 4.5%. She made no additional deposits and no w
Tresset [83]

Answer: 12 years

Step-by-step explanation:

6 0
3 years ago
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Does anyone know what 1+1 is?
hammer [34]

1+1 = 2  that is the answer


6 0
3 years ago
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Can someone check whether its correct or no? this is supposed to be the steps in integration by parts​
Gwar [14]

Answer:

\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

Step-by-step explanation:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

Given integral:

\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x

\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:

\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x

Using <u>integration by parts</u>:

\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

Therefore:

\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}

\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:

\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}

Therefore:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x

\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:

\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}

Divide both sides by 2:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}

Rewrite in the same format as the given integral:

\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

5 0
2 years ago
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