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2 years ago

the bottom of the swimming pool has an area of 48 meters. If the pool is 2 meters deep, what is the volume of the pool

Mathematics

1 answer:

KiRa [710]2 years ago

3 0

Volume of a rectangular prism=Base*Height
Area is 48 meters squared and height is 2 meters
V=48*2
V=96 cubic meters

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Common denominator of 5 6 7

Yanka [14]

Answer:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ... 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ... 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...

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2 years ago

Solve the equation.

Nutka1998 [239]

Answer:

Step-by-step explanation:

In this technique, if we have to factorise an expression like ax2+bx+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅−12=−12

AND

N1+N2=b=−1

After trying out a few numbers we get N1=3 and N2=−4

3⋅−4=−12, and 3+(−4)=−1

x2−x−12=x2−4x+3x−12

x(x−4)+3(x−4)=0

(x+3)(x−4)=0

Now we equate the factors to zero.

x+3=0,x=−3

x−4=0,x=4

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2 years ago

What is mAngleMHJ? 35 50 72.5 92.5

serg [7]

Answer:

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2 years ago

Please respond to the question below in your own words with proper explanations

ycow [4]

Answer:

$t(400)=673$ .

It means that 673 of 9 inch tiles are required to cover an area of 400 square feet, when spacing the tiles a quarter inch apart.

Step-by-step explanation:

Given:

The function relating number of 9 inch tiles required and area to cover is:

$t(A)=\frac{144}{9.25^2}A$

Now, plug in 400 for $A$ to evaluate $t(400)$ .

$t(400)=\frac{144}{9.25^2}\times 400=\frac{144\times 400}{85.5625}=\frac{57600}{85.5625}=673.19\approx6225$

Therefore, Don required 673 of 9 inch tiles to cover an area of 400 square feet, when spacing the tiles a quarter inch apart.

4 0

2 years ago

If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

2 years ago