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3 years ago

Explain how solving 8+2 can help you solve 80+20.

2 answers:

8 0

First do 8 + 2 and then it's gen right so you do the same just that this time you add a 0 to a 10 like8+2=10 add a zero and get 100

katrin [286]3 years ago

5 0

Becuase 80 is the same as 8 but has an extra 0 at the end (which makes it 80) and same for 20 to 2. 8 + 2 = 10 (now add a zero at the end). The answer should be 100.

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The diameter of a circle is 22 miles. What is the circle's area? Use 3.14 for pi.. Single choice.

Sonbull [250]

Answer:

379.94 mi²

Step-by-step explanation:

The area of a circle can be figured out with this formula: πr² (3.14 is π here, as it says in the directions).

We know the diameter is 22 miles. Divide that by 2 to get a radius of 11 miles.

The equation now becomes

3.14(11²)

11 squared is 121.

3.14(121)

I recommend splitting the 121 into 100, 20, and 1.

(3.14 * 100) + (3.14 * 20) + (3.14 * 1)

Multiply 3.14 by 100, 20, and 1.

314 + 62.8 + 3.14

Add 314 and 62.8.

376.8 + 3.14

Add 376.8 and 3.14

379.94 mi² is the area of a circle with a 22-mile diameter.

6 0

3 years ago

Find the following percentage of 4.2% of 50

aleksley [76]

4.2% of 50 is 2.1 So 2.1 is your answer

3 0

3 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

3 years ago

Please help !!! 3x+8y x = 7 y = -2

Vinil7 [7]

Answer: 5

Step-by-step explanation:

you do each variable individually first, 3(7)=21, 8(-2) is -16, then you add them, 21-16=5 so that is your answer

4 0

3 years ago

Identify the asymptotes of y=1 over x -6 A. B. C. D.

Westkost [7]

Answer:

D. x = 0; y = -6

Step-by-step explanation:

There is a vertical asymptote at the value of x that makes the denominator zero. That value is x=0. This is sufficient to choose the correct answer.

When x gets large, the term 1/x nears zero, so the value of y nears -6. This is the horizontal asymptote.

vertical asymptote: x = 0; horizontal asymptote: y = -6

6 0

3 years ago