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4 years ago

During one week at Jackson School, 210 students buy milk and 196 students buy juice. How many drinks are sold that week?

Mathematics

2 answers:

Stella [2.4K]4 years ago

7 0

210+196 = 406

406 drinks are sold that week

eduard4 years ago

6 0

406 because 210+196=406

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average age of 15 students of iub 11years if teacher is also included average age becomes 13 years how old is teachers

Yuki888 [10]

Answer: the teacher is 43

Step-by-step explanation: if you take 11 and multiply it by 15 you get 165 if you take 208 and divide it by 16 you get 13.

so basically you subtract 208 from 165 to get 43

3 0

3 years ago

A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer

choli [55]

<h2>Answer:</h2>

The area of the top surface of the washer is: 160.14 square mm.

<h2>Step-by-step explanation:</h2>

The top of the surface is in the shape of a annulus with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r is the inner radius.

Here we have:

$R=10\ mm\\\\and\\\\r=7\ mm$

Hence, we have:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$

4 0

3 years ago

Read 2 more answers

Help Plss <br>How many solutions do y=5-2x and 4x+2y = 10 have?

tia_tia [17]

y=2x+5

2(2x+5)+4x=10

4x+10+4x=10

8x=20

So there's one solution, x=2.5

5 0

3 years ago

A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the se

stealth61 [152]

Answer:

L = 2*√2

w = √2

Step-by-step explanation:

Given:

A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the semicircle.

Find:

What are the dimensions of the rectangle with maximum area?

Solution:

- Let the length and width of the rectangle be L and w respectively.

- We know that Length L lie on the diameter base. So , L < 4 and the width w is less than 2 . w < 2.

- Using the Pythagorean Theorem, we relate the L with w using the radius r = 2 of the semicircle.

r^2 = (L/2)^2 + (w)^2

sqrt (4 - w^2 ) = L / 2

L = 2*sqrt (4 - w^2 ) L < 4 , w < 2

- The relation derived above is the constraint equation and the function is Area A which is function of both L and w as follows:

A ( L , w ) = L*w

- We substitute the constraint into our function A:

A ( w ) = 2*w*sqrt (4 - w^2 )

- Now we will find the critical points for width w for which A'(w) = 0

A'(w) = 2*sqrt (4 - w^2 ) - 2*w^2 / sqrt (4 - w^2 )

0 = [2*sqrt (4 - w^2 )*sqrt (4 - w^2 ) - 2*w^2] / sqrt (4 - w^2 )

0 = 2*(4 - w^2 ) - 2*w^2

0 = -4*w^2 + 8

8/4 = w^2

w = + sqrt ( 2 ) ..... 0 < w < 2

- From constraint equation we have:

L = 2*sqrt (4 - 2 )

L = 2*sqrt(2)

7 0

4 years ago

Scientists determined that Antarctica’s average winter temperature was -34.44 degrees Celsius. The difference between the temper

jeka57 [31]

The highest temperature ever recorded for Antarctica was 17.5 degrees Celsius at Esperanza Base on the Antarctic Peninsula on March 24th, 2015. I know because I searched it up on google. Hope this helped!

-TTL

7 0

3 years ago