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3 years ago

13

What is the slope of the graph in the picture

2 answers:

BaLLatris [955]3 years ago

6 0

Answer:

-2

or

-2/1

Explanation:

**Slope = rise/run**

By counting the distance between the points, from (0,3) to (1,1) , it went down 2 units and right by 1 unit

Tasya [4]3 years ago

6 0

Answer:

m= -2

Step-by-step explanation:

Pick two points and subtract the y coordinates, then subtract the x coordinates, (0,3) and (1,1) or (1,1) and (2,-1)

$m = \frac{y_{2_ } - y_{1} }{x_{2} -x_{1} } = \frac{1-3}{1-0} = \frac{-2}{1} = -2$ $\frac{-1-1}{2-1} = \frac{-2}{1} = -2$

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Directions: Calculate the area of a circle using 3.14x the radius

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

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 ]

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<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>

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

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

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<h3>problem 5</h3>

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Now, let's calculate area ~

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<h3>problem 6</h3>

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

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