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kogti [31]
3 years ago
13

Joe has a 28 foot long board. He needs to cut it into 24 equal length parts.How many feet should each section of the board be?

Mathematics
1 answer:
Ilya [14]3 years ago
6 0
28 divides by 24 is 1.1666667 which rounds to 1.17 <——answer
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What’s the correct answer for this?
Serga [27]

Answer:

(0,2)

Step-by-step explanation:

2:4 means one part is 2/(2+4)=1/3 of AB and the other part is 2/3 of AB

Add 1/3 of the distance from -2 to 4. (1/3)(4+2)=2. -2+2=0 The x coordinate is 0

Subtract 1/3 of the distance from 6 to -6, (1/3)6+6)=4 6-4=2 The y coordinate is 2

The point is (0,2)

5 0
3 years ago
What is the length of the diagonal of a 10 cm by 15 cm rectangle?
Ksivusya [100]

Answer:

18.02 cm

Step-by-step explanation:

A diagonal makes a rectangle into two right triangles. So if we use the Pythagorean theorem we can find the hypotenuse which is the diagonals.

Remember, because we know the length and width, we know the two sides.

10^2 + 15^2 = c^2

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18.027 ≈ c

4 0
3 years ago
Read 2 more answers
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
Answer?????????????????
ANTONII [103]
So the angle A on the first diamond corresponds with angle Q, angle B with angle S, angle C with angle R, and angle D with angle P. So if angle D correspond (equals) angle P then x+34=97 and if angle R corresponds with C then 3y-13=83. From there just do some basic algebra to find the x and y values. 
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3 years ago
-5,8 and 2,-6 what the slope
Degger [83]

Answer:

bederr

Step-by-step explanation:

baderdkd0ensdeussdk fmdikjxijod djieos-sxmsososjs0hsoisnsohsowb9sbnszolsis

ssls0ajsojsjwoiwksjksosjoksoajaonsisj

i got 5 points mfor free haha!!

7 0
3 years ago
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