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Vladimir79 [104]
3 years ago
14

HELP ME PLS I WILL MARK YOU AS BRAINLIEST!!

Mathematics
1 answer:
Andrej [43]3 years ago
6 0

Answer:

the first one is x=5 and then the second one is x=-2/3

Step-by-step explanation:

I hope this helps!

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AveGali [126]
Length x width = area
<span>but
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3 years ago
Compute the product of 645.99 and 0.125, and round to the nearest tenth
Burka [1]

Answer:

80.7 because you multiply

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A 7-pack of tickets to the zoo costs $78.61. What is the unit price?
vredina [299]

Answer:

$11.23

Step-by-step explanation:

To find the unit price divide the price by the product. For example, 12 ounces of juice costs $2.45. You divide $2.45 by 12 ounces. The unit price would be $0.20.

<em>I hope this helped you, if it did more than all the other answers that may be on your question, feel free to give me brainliest, I would really appreciate it.</em>

7 0
2 years ago
Read 2 more answers
Find x and y part 4 for these problems​
Dvinal [7]

Answer:

Step-by-step explanation:

3) Sin30 = 11/x

x = 11/Sin30 = 11/0.5

x = 22

Tan 30 = 11/y

y = 11/tan30 = 11/0.5774

y = 19.1

4) Sin30 = 6/x

x = 6/Sin30 = 6/0.5

x = 12

Tan 30 = 6/y

y = 6/tan30 = 6/0.5774

y = 10.39

5) Sin45 = 9√2/y

y = 9√2/Sin45 = 9√2/(√2/2) =

9√2 × 2/√2 = 18

x = 18

Tan 45 = 9√2/x

x = 9√2/Tan 45 = 9√2/1

x = 9√2

6)

Sin60 = 9/x

x = 9/Sin60 = 9/0.866

x = 10.39

Tan 60 = 9/y/2 = 18/y

1.7321 = 18/y

y = 18/1.7321

y = 10.39

6 0
3 years ago
Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

=8\ln \left(\sqrt{2}+1\right)-0

so,

L=8\ln \left(1+\sqrt{2}\right)..............Answer

5 0
2 years ago
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