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Sonja [21]
2 years ago
10

If "-3(y-10=9" what is the value of y-12

Mathematics
2 answers:
zvonat [6]2 years ago
8 0

Answer:

the correct answer is -5 since y=7

vagabundo [1.1K]2 years ago
5 0
18917-8- 1819-7 18918
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Directions: Calculate the area of a circle using 3.14x the radius
Leokris [45]

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 1 \div 2

\qquad \sf  \dashrightarrow \:r = 0.5 \:  \: yd

Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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