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3 years ago

The diagram shows a cuboid.

Mathematics

1 answer:

torisob [31]3 years ago

8 0

Answer:

length x width x 6 = 90

area of shaded face = 15

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Select the equation that represents the problem. Mr. Paterson bought 104 crayons for his class. the crayons came in packs of 8.

MakcuM [25]

Answer:

Step-by-step explanation:

4 0

3 years ago

5n-3=|n+9| HELP ME PLEASE

ehidna [41]

Answer:

n = 3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

X squared divided by -10 =2<br> what is

lesya [120]

Hey there!

The answer to your question is ± 12

To solve the equation x²/-10 = 2, first multiply 10 to both sides:

x² = 20

Then, take the square root from both sides:

x = ± 20

4 0

3 years ago

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the v

Svetllana [295]

Answer:

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$

Step-by-step explanation:

Given - The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

To find - Write a differential equation that describes the relationship.

Proof -

Rate of change of volume V with respect to time t is represented by $\frac{dV}{dt}$

Now,

Given that,

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

⇒ $\frac{dV}{dt}$ ∝ $\sqrt[3]{V}$

Now,

We know that, when we have to remove the Proportionality sign , we just put a constant sign.

Let k be any constant.

So,

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$

6 0

3 years ago

What are the solutions of 3x2 + 6x + 6 = 0?

AlladinOne [14]

The solutions or roots to this equation is found by solving for x. We can do this a couple ways either by FOIL or quad formula

3x^2+6x+6=0
3(x^2+2x+2)=0

x^2+2x+2=0 we cant FOIL this out so we use the quad formula

x= [(-b+\-sqrt(b^2-4ac))/2a]

x= -2+\- sqrt(4-4(1)(2))/2(1)
x= -1 +i , -1 - i

So we have complex roots since our quad formula returned a negative number. Whenever the quad formula answer is positive we have two roots/solutions, when it is zero we have one root/solution, and whenever it is negative we have Complex roots/solutions

Hope this helps. Any questions please just ask. Thank you.

7 0

3 years ago