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ryzh [129]
3 years ago
11

Nick makes $50 in 4 hours How much money will he make in 1 hour?​

Mathematics
2 answers:
Nady [450]3 years ago
6 0

Answer:

$12,5

Step-by-step explanation:

50/4=12,5$

brainliest pls

goldfiish [28.3K]3 years ago
3 0

Answer:

$12.5

Step-by-step explanation:

let x be the money he make in 1 hour

50 : 4 = x : 1

4x = 50

x = 12.5

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Exercises
lyudmila [28]

Answer:

minimum 380 student tickets must be sold.

Step-by-step explanation:

c) 500x7=3500

intended  =5400

5400-3500=1900

1900/5= 380

3 0
2 years ago
Solve the following system of equations.<br><br> 2x+y=3<br> x=2y-1
Anvisha [2.4K]

Answer:

Step-by-step explanation:

1,1

5 0
3 years ago
Read 2 more answers
Someone pls help me.
nydimaria [60]
I’m here to help.
For this question here you need to use the sine formula for the area of a triangle and figure out the area of the sector

I’ll take you through it step-by-step

Area of the arc:

We only have 68.9 degrees out of 360 and we need to use the formula for the area of a circle.
Formula of a circle = pi x radius squared

68.9/360 x pi x 86.1184= 51.78 (2dp)

Area of triangle:
As you can see we have no height so we must use the sine formula for the area of a triangle.
Formula= 1/2abSinC

You should end up with
0.5 x 9.28x 9.28 x Sin(68.9)= 40.17 (2dp)

Now since you want the area of the segment shaded, just find the difference between these two values.

51.78-40.17 = 11.61
Answer= 11.6cm squared

Hope this helped!
3 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
What is the answer to this question?
steposvetlana [31]

Answer:

48x - 40y +24

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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