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Gnoma [55]
3 years ago
15

The length of the ___ of the rectangular region approximates the length of the ___

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0

Answer:

The length of the <em><u>height</u></em> of the rectangular region approximates the length of the <em><u>radius</u></em><em><u>.</u></em>

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Mr Kendrick is trying to organize his year 7 class into rows for their class photograph. If Mr Kendrick wishes to organize the 2
inysia [295]
Just list the factors of 24. Those are the possible numbers of students that can be in each row.

1,2,3,4,6,8,12,24 are the possible row sizes.
5 0
3 years ago
What is the volume of the cube below?
vladimir2022 [97]

Answer:

wht cube

Step-by-step explanation:

7 0
3 years ago
If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c
horsena [70]

Answer:

Part 1) cos(A + B) = \frac{140}{221}

Part 2) cos(A - B) = \frac{153}{185}

Part 3) cos(A - B) = \frac{84}{85}

Part 4) cos(A + B) = -\frac{36}{85}

Part 5) cos(A - B) = \frac{63}{65}

Part 6) cos(A+ B) = -\frac{57}{185}

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

sin(A)=\frac{8}{17}

cos(B)=\frac{12}{13}

Determine cos (A+B)

we know that

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{8}{17})^2=1

cos^2(A)+\frac{64}{289}=1

cos^2(A)=1-\frac{64}{289}

cos^2(A)=\frac{225}{289}

cos(A)=\pm\frac{15}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{15}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{13})^2=1

sin^2(B)+\frac{144}{169}=1

sin^2(B)=1-\frac{144}{169}

sin^2(B)=\frac{25}{169}

sin(B)=\pm\frac{25}{169}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{5}{13}

step 3

Find cos(A+B)

substitute in the formula

cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}

cos(A + B) = \frac{180}{221}-\frac{40}{221}

cos(A + B) = \frac{140}{221}

Part 2) we have

sin(A)=\frac{3}{5}

cos(B)=\frac{12}{37}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{3}{5})^2=1

cos^2(A)+\frac{9}{25}=1

cos^2(A)=1-\frac{9}{25}

cos^2(A)=\frac{16}{25}

cos(A)=\pm\frac{4}{5}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{4}{5}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{12}{37})^2=1

sin^2(B)+\frac{144}{1,369}=1

sin^2(B)=1-\frac{144}{1,369}

sin^2(B)=\frac{1,225}{1,369}

sin(B)=\pm\frac{35}{37}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{35}{37}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}

cos(A - B) = \frac{48}{185}+\frac{105}{185}

cos(A - B) = \frac{153}{185}

Part 3) we have

sin(A)=\frac{15}{17}

cos(B)=\frac{3}{5}

Determine cos (A-B)

we know that

cos(A - B) = cos(A) cos(B)+sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A-B)

substitute in the formula

cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}

cos(A - B) = \frac{24}{85}+\frac{60}{85}

cos(A - B) = \frac{84}{85}

Part 4) we have

sin(A)=\frac{15}{17}        

cos(B)=\frac{3}{5}

Determine cos (A+B)

we know that    

cos(A + B) = cos(A) cos(B)-sin(A) sin(B)

step 1

Find the value of cos(A)

Remember that

cos^2(A)+sin^2(A)=1

substitute the given value

cos^2(A)+(\frac{15}{17})^2=1

cos^2(A)+\frac{225}{289}=1

cos^2(A)=1-\frac{225}{289}      

cos^2(A)=\frac{64}{289}

cos(A)=\pm\frac{8}{17}

The angle A belong to the I quadrant, the cosine is positive

cos(A)=\frac{8}{17}

step 2

Find the value of sin(B)

Remember that

cos^2(B)+sin^2(B)=1

substitute the given value

sin^2(B)+(\frac{3}{5})^2=1

sin^2(B)+\frac{9}{25}=1

sin^2(B)=1-\frac{9}{25}

sin^2(B)=\frac{16}{25}

sin(B)=\pm\frac{4}{5}

The angle B belong to the I quadrant, the sine is positive

sin(B)=\frac{4}{5}

step 3

Find cos(A+B)

substitute in the formula    

cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}

cos(A + B) = \frac{24}{85}-\frac{60}{85}

cos(A + B) = -\frac{36}{85}

Download odt
4 0
3 years ago
What is the standard equation of the circle in the graph
ivanzaharov [21]

Answer:

option 1

Step-by-step explanation:

formula:(x-h)^(2)*(y-k)^(2)=r^2

well if you move the circle 3 units left h is (-3), plug it in, you get 3, so it is (x+3)^2*(y-k)^(2)=r^2

if you move the circle 2 units up, plug it in you get (-2), so it is

(x+3)^2*(y-2)^(2)=r^2

now, the radius is 3 (we can see from the graph) so radius ^2 is 9, thus

(x+3)^2*(y-2)^(2)=9

this is the answer, and the option that has it is answer is option 1

6 0
3 years ago
By graphing the system of constraints, find the values of x and y that maximize the objective function.
Anastasy [175]
The easiest method to solve problems like this is to graph the inequalities given and shade the regions that make them true. When you have properly shaded all of the regions, you will find that you have a region which is bounded on all four sides by one of the inequalities, and then you can find the x and y values which correspond to the vertices of the shaded region.

You didn't provide a function that you are trying to maximize in this example, but the idea is that you take all of the (x,y) points which correspond to the vertices and plug them into your objective function. The one which produces the largest value maximizes it (it is a similar process for minimizing it, but you'd be looking for the smallest value). Let me know if you need more help than that, or would like me to work out the example you have provided (I will need an objective function for it though).
6 0
2 years ago
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