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3 years ago

Mary and Julie both have a goal of raising $1000 for charity. Mary raises $12 every 4 days. Julie raises $19 every 6 days.

2 answers:

9 0

Mary would have a rise of 12 and a run of 4, making her slope, or rate of change, $3 a day. Julie would have a rise of 19 and a slope of 6, making her slope, or rate of change, $3.17 a day. Since Julie is raising money at a faster rate, she would get to the goal of $1000 first.

Math Genius

2 years ago

Thank you! :3

baherus [9]3 years ago

6 0

Julie would reach her goal first. Julie would reach her goal in approximately 316 days while it would take Mary approximately 333 days to reach it.

I know this because the steps to solve the problem are:

1.) 1,000/12 = 83.333 (to find how many days it would take her everyday)

2.) 83.333x4 = 333 (multiplying the answer by four makes it every four days)

3.) determine that it takes Mary 333 days

4.) 1,000/19 = 52.631 (just like step 1 but with different numbers)

5.) 52.631x6 = 316 (just like in step 2 but different numbers)

6.) determine it takes Julie 316 days

7.) 316 < 333 (takes Julie less time)

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Find the Slope given the points: (1,8) and (-2,-1)

Karolina [17]

Answer:

m = 3

Step-by-step explanation:

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3 years ago

A _______ is the numerical factor of a multipication expression like 4x

Sergio039 [100]

I think it is coefficient, hope this helps!

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3 years ago

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0$

The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

8 0

3 years ago

-8(8 - x) = 4/5(x + 10)

Vedmedyk [2.9K]

Answer:

x = 10

Step-by-step explanation:

Solve for x:

-8 (8 - x) = (4 (x + 10))/5

Multiply both sides by 5:

-8×5 (8 - x) = (5×4 (x + 10))/5

(5×4 (x + 10))/5 = 5/5×4 (x + 10) = 4 (x + 10):

-8×5 (8 - x) = 4 (x + 10)

5 (-8) = -40:

-40 (8 - x) = 4 (x + 10)

Expand out terms of the left hand side:

40 x - 320 = 4 (x + 10)

Expand out terms of the right hand side:

40 x - 320 = 4 x + 40

Subtract 4 x from both sides:

(40 x - 4 x) - 320 = (4 x - 4 x) + 40

40 x - 4 x = 36 x:

36 x - 320 = (4 x - 4 x) + 40

4 x - 4 x = 0:

36 x - 320 = 40

Add 320 to both sides:

36 x + (320 - 320) = 320 + 40

320 - 320 = 0:

36 x = 40 + 320

40 + 320 = 360:

36 x = 360

Divide both sides of 36 x = 360 by 36:

(36 x)/36 = 360/36

36/36 = 1:

x = 360/36

The gcd of 360 and 36 is 36, so 360/36 = (36×10)/(36×1) = 36/36×10 = 10:

Answer: x = 10

7 0

3 years ago

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

5 0

2 years ago