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3 years ago

Need help with this!!

Mathematics

1 answer:

Taya2010 [7]3 years ago

8 0

Answer:

The answer is B..... The corresponding is the identical number between all

The is 7 in each...

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12. The $60 selling price of a camera is the cost

lbvjy [14]

Answer:

I can help but your question does not make sense. You said the price of the camera is $60 and the question is find the cost?

Step-by-step explanation:

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4 years ago

whe value of two rational number is 2,if one of the rational number is 1/7 so what is the value of other brainly

baherus [9]

Answer:

13/7

Step-by-step explanation:

X + 1/7 = 2

X = 2- 1/7

x = 13/7

6 0

3 years ago

Melissa and Tomas are playing a game with complex numbers. If Melissa has a score of 5-4i and Tomas has a score of 3+2i, what is

Gemiola [76]

8-2i is their total score. You can't combine the imaginary numbers and real numbers.

3 0

3 years ago

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The object of the game is to toss a beanbag in the circular hole of a 48-by-24-inch board. If the diameter of the circle is 6 in

larisa [96]

Answer:

Geometric probability of an object hitting a circular hole is 0.0245.

Step-by-step explanation:

We have given,

A board of size 48 by 24 inch. There is a circular hole in the board having diameter 6 inches.

So,

Area of a board = 48 × 24 = 1152 square inches

And area of circular hole = π×r² {where r = diameter/2 = 6 / 2 = 3 inches}

Area of circular hole = π×3² = 9π = 28.27 square inches

Now, we need to find the geometric probability of an object will hit the circle.

Geometric probability = Area of circular hole / Area of board

Geometric probability = $\frac{28.27}{1152}$

Geometric probability = 0.0245

hence geometric probability of an object hitting a circular hole is 0.0245.

6 0

3 years ago

Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?

Mademuasel [1]

Answer:- a.The given expression is equivalent to $\frac{x^{10}y^{14}}{729}$

Given expression:- $[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}$

$=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}$

$=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}$

$=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}$

$=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}$

$=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}$

Thus a. is the right answer.

6 0

3 years ago

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