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oee [108]
3 years ago
7

Use Long Division to answer number 5. Please show all work!​

Mathematics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

Step-by-step explanation:

The steps for the long division method are shown in the attached photo.

The quotient is 3x² + 3x - 3 and the remainder is 30x - 2

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Ronald is trying to reconstruct his spending pattern from May. He knows that he had $315 in his account on May 1, but after that
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Answer: C. Ronaldo overviewed his account twice.

Step-by-step explanation:

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7 0
3 years ago
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find the length of each isosceles right triangle when the hypotenuse of the given measure given 5 to the square root of 6
iragen [17]

In isosceles right triangle, legs are equal. Let each leg is of x units . And the given hypotenuse is 5 square 6. We use pythagorean identity here, which is

a^2 + b^2 = c^2

Where a=b=x and c= 5 square root 6

x^2 +x^2 =(5 \sqrt6)^2

2x^2 = 150

x^2 =75

x= 5 \sqrt 3

And that's the length of both the legs . ANd that's the required answer .

8 0
3 years ago
HELP GEOMTRY TEST !!!
Inessa05 [86]

Answer:

B

Step-by-step explanation

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5 0
3 years ago
There are 250 bricks used to build a wall 20 feet high.How many bricks will be used to buld a wall that is 30 fet high
Dmitry_Shevchenko [17]

Answer: 375

Step-by-step explanation:

Since we are given the information that there are 250 bricks used to build a wall 20 feet high, the number of brick used per feet will be:

= 250/20

= 12.5 bricks per feet.

Therefore, the number of bricks that will be used to buld a wall that is 30 fet high will be:

= 12.5 × 30

= 375 bricks.

Therefore, 375 bricks would be used.

3 0
3 years ago
Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif
xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0
3 years ago
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