Ask question

Ask question

All categories

3 years ago

8

Z the weight of an object on Mars varies directly with its weight on Earth an object that weighs 50 pounds on Mars weigh 150 pou

nds on Earth if an object weighs 120 pounds on Earth write and solve a direct variation equation to find out how much an object would weigh on Mars

1 answer:

a_sh-v [17]3 years ago

3 0

If 150 ibs on Earth is 50ibs on Mars then 120 on Earth would be 20 ibs on Mars.

You might be interested in

Slope = four fifths ,y-intercept = -1

Nadya [2.5K]

Slope-intercept form:
y=mx+b

m=slope
b=y-intercept

Data:
m=4/5
b=-1

y=mx+b
y=(4/5)x-1

Answer: the equation of the line with slope=4/5, and y-intercept=-1 would be:
y=(4/5)x-1

5 0

3 years ago

5 Exam-style ABCD is a kite.

Mnenie [13.5K]

let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

$y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line whose slope is -1/3 and passes through point A

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}$

5 0

2 years ago

What is $35 of $175 please

Karo-lina-s [1.5K]

If your looking for the percent you do 35 × 100, then divide by 175. You should get 20% as your answer.

7 0

2 years ago

Read 2 more answers

Can someone help me find the volume of this?

GalinKa [24]

To find the volume of any cube you need to know the length, width and height. The formula to find the volume multiplies the length by the width by the height. The good news for a cube is that the measure of each of these dimensions is exactly the same. Therefore, you can multiply the length of any side three times.

5 0

3 years ago

What is the coordinate?

faltersainse [42]

Remember that the equation of a circle is:
$(x - h)^{2} + (y - k)^{2} = r^{2}$
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.

$x^{2} - 6x + y^{2} - 10y = 56$

Complete the square. We must do this for x² - 6x and y² - 10y separately.

x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9

Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.

$x^{2} - 6x + 9 + y^{2} - 10y = 65$

Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.

y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.

$(x - 3)^{2}+ y^{2} - 10y + 25= 90$

Factor y² - 10y + 25 to get (y - 5)²

$(x - 3)^{2}+ (y - 5)^{2} = 90$

Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)

Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.

7 0

2 years ago