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2 years ago

The volume of the prism is 840 cubic units. What is the perimeter of the base?

Mathematics

2 answers:

Masteriza [31]2 years ago

4 0

Answer:

30 Units

Step-by-step explanation:

eimsori [14]2 years ago

3 0

Same as what I explained in your last question.

840, base, perimeter.

Hope this helps!!!

Thanks for using Brainly!

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Which equation has an a-value of –2, a b-value of 1, and a c-value of 3? 0 = –2x2 + x + 3 0 = 2x2 + x + 3 0 = –2x2 + 3 0 = 2x2 –

Zolol [24]

ax² + bx + c = 0

You know:

a = -2

b = 1

c = 3

The 1st option is your answer

-2x² + x + 3 = 0

a = -2

b = 1

c = 3

6 0

2 years ago

The quotient of 38 times a number and -4

nikdorinn [45]

Answer:

140y

Step-by-step explanation:

35 times a number times -4 is 35×y×(-4) which is -140y

4 0

1 year ago

3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses.

Rufina [12.5K]

Answer:

a) The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 , P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

8 0

2 years ago

A distributor buys packs of trading cards from a wholesaler for $0.80 each. The distributor marks up the cards by 100% before se

Llana [10]

Answer:

$1.60

$2.88

Step-by-step explanation:

To find the selling price of the distributor and the retailer, we first need to find how much the distributor sells each pack of cards.

To find the selling price we use the formula.

Selling Price = Cost + Markup

Cost = $0.80

Markup rate = 100% or 1

Selling Price = 0.80 + (0.80*1)

Selling Price = $1.60

So the distributor sells each pack of cards at $1.60 to the retailer.

Now to find the selling of the retailer, we need to use the selling price of the distributor.

Cost = $1.60

Markup rate = 80% or 0.80

Selling price = 1.60 + (1.60 * 0.80)

Selling price = $2.88

So the retailer sells each pack of cards at $2.88 to the customers.

8 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago