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Maksim231197 [3]
3 years ago
13

Find the area of the shaded sector. Leave your answer in terms of

Mathematics
1 answer:
lina2011 [118]3 years ago
8 0
Area is (90/360)(pi)(64)= 16pi or 50.24
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Simplify<br> a(cube)-1000b(cube)<br> 64a(cube)-125b(cube)
Elanso [62]

The simplification of a³ - 1000b³ and 64a³ - 125b³ is (a - 10b) × (a² + 10ab + 100b²) and 4a - 5b) • (16a² + 20ab + 25b²) respectively.

<h3>Simplification</h3>

Question 1: a³ - 1000b³

a³ - b³

= (a-b) × (a² +ab +b²)

  • 1000 is the cube of 10
  • a³ is the cube of a¹
  • b³ is the cube of b¹

So,

(a - 10b) × (a² + 10ab + 100b²)

Question 2: 64a³ - 125b³

a³ - b³

= (a-b) × (a² +ab +b²)

  • 64 is the cube of 4
  • 125 is the cube of 5
  • a³ is the cube of a¹
  • b³ is the cube of b¹

So,

(4a - 5b) • (16a² + 20ab + 25b²)

Learn more about simplification:

brainly.com/question/723406

#SPJ1

8 0
2 years ago
How to find the constant of proportionality on a graph
Akimi4 [234]
Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate.
4 0
3 years ago
Sweet noticed the ratio of long boards to short boards was 1:4. If there were 35 surfboards at the beach , how many were short b
andre [41]
So the ratio to longboard to short board is 1:4, the total of that ratio is 5. 35 ÷ 5 = 7. now we're going to take those ratios and multiply 1 × 7 = 7 and 4 × 7 = 28. Your final answer is B. to confirm your answer add 28 + 7.
7 0
3 years ago
Are the figures below similar? Why or why not? Determine whether the triangles shown in the image are similar.
Vika [28.1K]

Answer:

B) no, because the corresponding angles are not congruent

Step-by-step explanation:

Similar triangles must have proportional sides, as well as congruent corresponding angles. In this instance, we can see that the angles are not <em>congruent</em>, and so there is no need to solve for proportion.

~

3 0
3 years ago
Read 2 more answers
Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)
Dima020 [189]

Split C into two component segments, C_1 and C_2, parameterized by

\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)

\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)

respectively, with 0\le t\le1, where \mathbf r_i(t)=(x(t),y(t)).

We have

\mathrm d\mathbf r_1=(6,1)\,\mathrm dt

\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt

where \mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt

so the line integral becomes

\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)

=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt

=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6

6 0
3 years ago
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