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3 years ago

The square ABCD is divided into eight equal parts. The shaded area is 25 cm². What is the area of the square ABCD

Mathematics

1 answer:

Akimi4 [234]3 years ago

6 0

Answer:

I think the answer is 200cm^2

Step-by-step explanation:

since the square is divided into eight equal parts

and one shaded part is =25cm^2

multiply the area of the shaded part by the number of equal parts

= 25cm^2 ×8

= 200cm^2

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For the linear equation, find the product of 5 and the linear equation, and solve both equations for y. 2x + 7y = 11

tensa zangetsu [6.8K]

Answer:

(a) $10x + 35y = 55$

(b) $y = \frac{11 - 2x}{7}$

Step-by-step explanation:

Given

$2x + 7y = 11$

Solving (a): Multiply equation by 5.

Multiply both sides by 5

$5 * (2x + 7y) = 11*5$

Open bracket

$5 * 2x + 5 * 7y = 11*5$

$10x + 35y = 55$

(b) Solve for y in $10x + 35y = 55$ and $2x + 7y = 11$

$10x + 35y = 55$

Subtract 10x from both sides

$35y = 55 - 10x$

Divide both sides by 35

$\frac{35y}{35} = \frac{55 - 10x}{35}$

$y = \frac{55 - 10x}{35}$

Factorize the numerator:

$y = \frac{5(11 - 2x)}{35}$

Express 35 as 7 * 5

$y = \frac{5(11 - 2x)}{7 * 5}$

$y = \frac{(11 - 2x)}{7}$

$y = \frac{11 - 2x}{7}$

When $2x + 7y = 11$ is solved, the solution will be: $y = \frac{11 - 2x}{7}$

6 0

3 years ago

Please someone help ASAP what is the correct answer ?

kramer

Answer:

It is x2 -4 because it correct

6 0

3 years ago

If the probability of being hospitalized during a year is 0.15 find the probability that no one in the family of 4 will be hospi

Goryan [66]

The probability that no one in the family of 4 will be hospitalized during a year is 0.52

Step-by-step explanation:

The probability for a person to be hospitalized during one year is

$p(h)=0.15$

As a consequence, the probability for a person of NOT being hospitalized during one year is

$p(h^{-1})=1-0.15=0.85$

If we have 4 people in the family, the probability of each of them of NOT being hospitalized during the year is independent from the probability of the others - therefore, these are independent events. This means that the total probability of the 4 people of not being hospitalized during a year is:

$p(h^{-1})^4=(0.85)^4=0.52$