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4 years ago

I am doing tessellations I need help what does this look like

Mathematics

1 answer:

Naily [24]4 years ago

7 0

I can’t do it rn so just dm me later

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The picture shows a circular clock face.

marysya [2.9K]

Answer:

[C] 25π square inches

Step-by-step explanation:

Given that:

the long hand of the clock is about 5 inches long.

To Find:

What is the approximate area of the clock face?

Solve:

Formula - A =πr²

Note that;

π = 3.14 (about)

Radius - 5 inches

A =πr²

A = 3.14(5)²

A = 3.14(25)

A = 78.5

Now let see the answer choices:

A. 5π square inches ≈ 5(3.14) = 15.7

B. 10 π square inches ≈ 10(3.14) = 31.4

C. 25 π square inches ≈ 25(3.14) = 78.5

D. 100 π square inches ≈ 100(3.14) = 314

Hence, the answer is [C] 25 π square inches

Kavinsky~

6 0

2 years ago

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Is y=-½x+2 perpendicular to -2x+1

BartSMP [9]

Answer:

i think no sorry if im whrong

3 0

3 years ago

What is 7 over 10 as a decimal

Inessa05 [86]

.70 yup that's it that's the answer

4 0

3 years ago

F(t)=t^2+19t+60 What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$