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3 years ago

What's the area of the Square 28in• 28in

Mathematics

2 answers:

serg [7]3 years ago

7 0

The area is 784 inches

N76 [4]3 years ago

5 0

28in x 28in is 784in
For example used distributive property of multiplication

28 x 28= (28 x 20) + (28 x 8)

= (28 x 20) + (28 x 8)

=560 + 224

=784
;)

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Answer and explanation please

Shalnov [3]

The Smallest possible value for (2 × a) - (a × b) if Positive integers a and b are each less than 6 is 1

<h3>What are positive integers?</h3>

Positive integers refers to whole numbers, that are greater than zero and do not include fractions or decimals. They include; 1, 2, 3, 4, 5, 6, 7, 8 etc

Positive integers a and b are each less than 6

Positive integers less than 6 are 1, 2, 3, 4, 5

Smallest possible value for (2 × a) - (a × b)

So, 1 is the smallest positive integer less than 6

(2 × a) - (a × b)

= (2 × 1) - (1 × 1)

= 2 - 1

= 1

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What is the solution of the equation: -3m - 21 =-6 answer choices -45 -5 9 -81

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Step-by-step explanation:

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4 years ago

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Jennifer had $30 to spend on herself. She spent 1/5 of the money on a sandwich, 1/6 for a ticket at the museum, and 1/2 of it on

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Answer:

Step-by-step explanation:

because 1/5 of 30 is 5

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4 years ago

I need help on two questions!

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1) Answer: $\frac{x^{2}}{100} +\frac{y^{2}}{4} = 1$

Explanation:

The equation of an ellipse is: $\frac{(x-h)^{2}}{a^{2}} +\frac{(y-k)^{2}}{b^{2}} = 1$ ; where (h, k) is the center, "a" is the x-radius, and "b" is the y-radius.

Center Radius

x-axis: (10 + -10)/2 = 0 10 - 0 = 10

h = 0 a = 10

y-axis: (2 + -2)/2 = 0 2 - 0 = 2

k = 0 b = 2

Now, input the values into the equation:

$\frac{(x-0)^{2}}{10^{2}} +\frac{(y-0)^{2}}{2^{2}} = 1$

$\frac{x^{2}}{100} +\frac{y^{2}}{4} = 1$

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2) Answer: $\frac{x^{2}}{1} +\frac{(y+2)^{2}}{4} = 1$

Explanation:

Vertices are: (0, 1) and (0, -5) ------> x-values are the same, y = 1, -5

Covertices are: (-1, -2) and (1, -2) ----> y-values are the same, x = -1, 1

Center Radius

x-axis: (-1 + 1)/2 = 0 1 - 0 = 1

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y-axis: (1 + -5)/2 = -2 1 - (-2) = 3

k = -2 b = 3

Now, input the values into the equation:

$\frac{(x-h)^{2}}{a^{2}} +\frac{(y-k)^{2}}{b^{2}} = 1$

$\frac{(x-0)^{2}}{1^{2}} +\frac{(y-(-2))^{2}}{3^{2}} = 1$

$\frac{x^{2}}{1} +\frac{(y+2)^{2}}{9} = 1$

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4 years ago