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masha68 [24]
3 years ago
6

What’s all the prime numbers between 20 and 40

Mathematics
2 answers:
adoni [48]3 years ago
7 0

Step-by-step explanation:

All the prime numbers between 20 and 40 are ⤵

23, 29, 31 , 37

Hope it will help :)

Ad libitum [116K]3 years ago
5 0

Answer:

Step-by-step explanation:

23, 29, 31, 37.

:))

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A circle has a diameter of 24 units. Find the circumference and area of the circle in terms of π.
enyata [817]
Circumference=24 pi
Area=144 pi, shown clearly in photo

7 0
3 years ago
Read 2 more answers
PLEASE HELP ILL GIVE BRAINLIEST
gogolik [260]

Answer:

3/4 i believe

Step-by-step explanation:

4 0
3 years ago
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What is the area, in square inches, of the isosceles trapezoid below?
zimovet [89]

Answer:

104.8 in^2

Step-by-step explanation:

There are 2 ways to solve this problem.

The 1st way:

Let's make 2 triangles and 1 rectangle:

Rectangle Length = 8.3

Rectangle Width = 8

So, the left out length will be 17.9 - 8.3

=> 9.6

Since, 9.6 cm is for 2 parts.

=> 9.6 / 2

=> 4.8

So, Height of the Triangle = 8

      Base of the triangle = 4.8

Area of a rectangle

=> 8.3 x 8

=> 66.4

Area of the triangle

=> 1/2 x 8 x 4.8

=> 4 x 4.8

=> 19.2

There are 2 triangles:

=> 19.2 x 2

=> 38.4

=> 66.4 + 38.4

=> 104.8

The area of the trapezoid = 104.8 in^2.

The 2nd way is:

Area of a trapezoid

=> Smaller Base + Larger Base / 2 x Height

=> 8.3 + 17.9 / 2 x 8

=> 26.2 / 2 x 8

=> 13.1 x 8

=> 104.8

The area of the trapezoid is 104.8 in^2

6 0
3 years ago
Its due today please help!
Korolek [52]

Answer:

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5 0
3 years ago
Find the approximate area between the curve f(x) = -4x² + 32x and on the x-axis on the interval [0,8] using 4 rectangles. Use th
Doss [256]

Split up the interval [0, 8] into 4 equally spaced subintervals:

[0, 2], [2, 4], [4, 6], [6, 8]

Take the right endpoints, which form the arithmetic sequence

r_i=2+\dfrac{8-0}4(i-1)=2i

where 1 ≤ <em>i</em> ≤ 4.

Find the values of the function at these endpoints:

f(r_i)=-4{r_i}^2+32r_i=-16i^2+64i

The area is given approximately by the Riemann sum,

\displaystyle\int_0^8f(x)\,\mathrm dx\approx\sum_{i=1}^4f(r_i)\Delta x_i

where \Delta x_i=\frac{8-0}4=2; so the area is approximately

\displaystyle2\sum_{i=1}^4(-16i^2+64i)=-32\sum_{i=1}^4i^2+128\sum_{i=1}^4i=-32\cdot\frac{4\cdot5\cdot9}6+128\cdot\frac{4\cdot5}2=\boxed{320}

where we use the formulas,

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

6 0
3 years ago
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