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ahrayia [7]
3 years ago
13

Find the highest odd number that is a factor of 40 and a factor of 60.

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

HCF (HIGHEST COMMON FACTOR) OR GCF ( GREATEST COMMON FACTOR) OF 40 AND 60 IS 20.

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christen has to run around the rectangle soccer field during practice. The soccer field measures 60 meters long by 45 meters wid
Elena L [17]

Answer:

210 m

Step-by-step explanation:

Perimeter is the distance around the outside of a two-dimensional shape.

Area is the space inside of a two-dimensional shape. We can also think of area as the amount of space a shape covers.

Christen ran around the soccer field, so she ran around the perimeter.

We can find the perimeter by adding all of the side lengths.

Perimeter=60 m+45 m+60 m+45 m

Christen ran 210m.

7 0
3 years ago
Use the method of least squares to solve the following problem.
slamgirl [31]
Purr fbfbfbffjhfbfbfbfbfbfbf
3 0
3 years ago
4 1/2 divided by 2 3/8
Yuki888 [10]
First you make them mixed numbers:
9/2 ÷ 19/8
Then you flip the second fraction around
9/2 ÷ 8/19
Then you change ÷ to ×
9/2 × 8/19
And then multiply across!
72/38 = 1 34/38 = 1 17/19 (the final answer)
Hope this helps!
3 0
3 years ago
Dodd knit
Alecsey [184]

Answer:

108

Step-by-step explanation:

The total of stitches in width is 108, if 9 inches is the total inches in width. The remaining stitches of the remaining width (if this is the case) would be 84 stitches. If we are combining 9 and 2, the total amount of stitches would be 132. If you would like the work, you can ask otherwise hope this helped!

5 0
3 years ago
Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt
sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

3 0
3 years ago
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