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3 years ago

The shape above has the following

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

8 0

It’s A 44-1) and if not then it’s b

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25 pts and a mark for real answers. Help!! A store is having a sale on chocolate chips and walnuts. For 5 pounds of chocolate ch

kozerog [31]

Answer:

Each pound of chocolate chips costs $ 1.5 and each pound of walnuts cost $ 3.75 .

Step-by-step explanation:

Let, each pound of chocolate chips costs $ x and each pound of walnuts costs $ y.

Then, according to the question,

5x + 6y = 30 --------------(1) and,

3x + 2y = 12 ----------------(2)

Multiplying (2) by 3 we get,

9x + 6y = 36 ---------------------(3)

Deducting (1) from (3) we get,

4x = 6

⇒ x = 1.5 ----------------------(4)

From (4), putting the value of x in (2) , we get,

2y = 7.5

⇒ y = 3.75 ---------------------------(5)

3 0

3 years ago

Find the slope and the y-intercept of the line

Rom4ik [11]

Answer:

the slope of the line is 4

m = 4

y-intercept is -8

y = 4(0) - 8 = 0 - 8 = -8

3 0

4 years ago

What is 20/10 as a mixed number?

nadya68 [22]

Hello there!
the answer is 2

Hope this helps! :)
~zain

8 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

Mikayla withdraws $20 on 4 different days durin the week.Find the total change in her account balance after the withdrawls.

Stells [14]

Wait what did she have in her account? We need to know to answer.

7 0

3 years ago