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3 years ago

PLEASE HELP!! (will give brainliest for best answer)

Mathematics

1 answer:

Likurg_2 [28]3 years ago

5 0

Answer:

i think the answer should be 60 ft long bc that would leave 10ft on each side of the banner

Step-by-step explanation:

hope this helps

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There were 276 people on air airplane write a number greater than 276

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277 good luck on whatever you need this for

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3 years ago

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I need help with this problem

-BARSIC- [3]

Regular is
$97.99 after sale price is 83.30
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3 years ago

Determine whether quadrilateral ABCD with vertices A(–4, –5), B(–3, 0), C(0, 2), and D(5, 1) is a trapezoid. Step 1: Find the sl

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Answer:

The Slope of AB: 5/1

The Slope of DC: -1/5

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The Slope of AD: 2/3

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Step-by-step explanation:

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Integrate e^x(sin(x) cos(x))

Karo-lina-s [1.5K]

$I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx$

$\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}$

Looking for $I_2$ :

$\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I$

Replacing:

$2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}$

6 0

3 years ago

In the diagram, TC represents a vertical building. The points, A and B, are on the same level as the foot C of the building such

Alenkinab [10]

Answer:

(a) The height of the building is 60.06 m

(b) The distance AB is 139.43 m

Step-by-step explanation:

The given parameters are

Given that segment BT = segment AT + 29

By trigonometric ratios, we have;

cos∠ATC = CT/AT

cos∠BTC = CT/BT

Therefore, we have;

cos(40°) = CT/AT.................................(1)

cos(56°) = CT/BT = CT/(AT + 29).....(2)

cos(56°) = CT/(AT + 29)......................(3)

From equation (1)

CT = AT×cos(40°)

From equation (3)

AT×cos(56°) + 29 × cos(56°) = CT

Therefore;

AT×cos(40°) = AT×cos(56°) + 29 × cos(56°)

AT×cos(40°) - AT×cos(56°) = 29 × cos(56°)

AT×(cos(40°) - cos(56°)) = 29 × cos(56°)

AT = 29 × cos(56°)/(cos(40°) - cos(56°)) = 78.4 m

TC = CT = AT×cos(40°) = 78.4×cos(40°) = 60.06 m

The height of the building = 60.06 m

(b) BT = AT + 29 = 78.4 m + 29 m= 107.4 m

AB = AT×sin(∠ATC ) + BT×sin(∠BTC) = 78.4×sin(40°) + 107.4×sin(56°) = 139.43 m

The distance AB = 139.43 m.

6 0

3 years ago