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3 years ago

12

On Earth you weigh 54.5 kilograms, while on the moon you weigh 9,071 grams. How many more grams do you weigh on Earth than on th

e moon?

2 answers:

Nana76 [90]3 years ago

8 0

Answer:

Earth weight:

54.5 kilograms = 54.5 × 10 × 10 × 10 = 54,500 grams

Difference of weights:

54,500 g − 9,071 g = 45,429 grams

Step-by-step explanation:

Sever21 [200]3 years ago

4 0

Answer:

45,429

Step-by-step explanation:

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A right triangle has a base of 11 yards and a height of 7 yards . If you were to construct a similar but not congruent right tri

Dimas [21]

Answer:

bae of 44 yds and a height of 28 yds

Step-by-step explanation:

11x7 = 88 yd^2 = area of small triangle

616/88 = 16 so the larger triangle area is 16 times larger than the smaller triangle.

The area of a triangle is not proportional to it's dimensions. An increase of 16 times the area only increases the dimensions by 4 times.

larger triangle dimensions are 11 x 4 = 44 and 7 x 4 = 28

5 0

3 years ago

Determine the mean of the numbers 6, 17, 22, 28, and 37.

Kipish [7]

To find the mean or the avg you add up all the numbers you have and then divide by how many numbers you added together in this case 5

$6+ 17+22+28+37=110$

$\frac{110}{5} =22$

so the mean is 22, hope this helped you!

4 0

3 years ago

Help ya’ll smart people help me

torisob [31]

Answer:

1) -200

2) -126

3) -5 + (-10)

4) -10

5) -60 + (-85) + (-115) = -260

Step-by-step explanation:

You can just add all of them normally then add the negative sign (-) at the end.

Hope this helps!

3 0

3 years ago

Any help if possible Thank you ❤

azamat

I believe it's 37CM, I had that question on my finals last year.

8 0

3 years ago

Read 2 more answers

1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Sveta_85 [38]

Answer:

1. $-5x+3y+44=0$

2. $2x+y-2=0$

3. $2x+y-4=0$

Step-by-step explanation:

Standard form of a line is $Ax+By+C=0$ .

If a line passing through two points then the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope, i.e., $m=\dfrac{y_2-y_1}{x_2-x_1}$ .

1.

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

$y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)$

$y+3=\dfrac{-5}{-3}(x-7)$

$y+3=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$-5x+3y+9+35=0$

$-5x+3y+44=0$

Therefore, the required equation is $-5x+3y+44=0$ .

2.

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to $2 x + y =-5$ .

Slope of the line : $m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2$

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

$y-2=-2(x-0)$

$y-2=-2x$

$2x+y-2=0$

Therefore, the required equation is $2x+y-2=0$ .

3.

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to $2x + y =-5$ .

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

$y-0=-2(x-2)$

$y=-2x+4$

$2x+y-4=0$

Therefore, the required equation is $2x+y-4=0$ .

5 0

2 years ago