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2 years ago

Can y’all help me plz?

Mathematics

1 answer:

gulaghasi [49]2 years ago

3 0

Answer:

Step-by-step explanation:

Pythagreom Therom:

a^2+b^2=c^2

12^2+5^2=c^2

144+25=c^2

169=c^2

c=13

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zavuch27 [327]

Answer:

32 feet high

Step-by-step explanation:

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2 years ago

The graph shows the amount of money paid when purchasing bags of peanuts at the zoo:

abruzzese [7]

Answer:

dive the cost by the number of bags for each dot:

4 / 1 = 4

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12 / 3 = 4

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Step-by-step explanation:

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3 years ago

State the domain of the numbers displayed in column x and column y. x y

ella [17]

We can not really tell in this question as you dont know the equation that is being used for the domain and range relationship but overall one should know that:

The set of values of the independent variable(s) for which a function or relation is defined as the domain of a function. Typically, this is the set of x-values that give rise to real y-values.

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3 years ago

Rewrite using positive exponents

Oxana [17]

Answer:

Your selection is appropriate

Step-by-step explanation:

A negative exponent in the numerator is equivalent to a positive exponent in the denominator, and vice versa.

... a⁻² = 1/a²

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3 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$