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anyanavicka [17]

3 years ago

11

Randy and Jason have $5.00 each after they pay for their tickets. The shared a big popcorn for $2.50 and a jumbo soda for $3.50.

How much money does each boy have left?

1 answer:

Alik [6]3 years ago

6 0

Answer:

they have 3 dollars left because you add 3.50 plus 2.50 equals 6 then you divide it by 2 cuz 2 people are sharing and they have 5 bucks each which would get you the answer of 3. they have 3 dollars each

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Solve the system shown.

krek1111 [17]

1. Rewrite the system:

x+12y=68 (i)
x=8y-12 (ii)

2. Let's substitute the equation (ii) into the equation (i):

x+12y=68
(8y-12)+12y=68

3. Then, you have:

8y-12+12y=68

4. When you clear "y", you have:

20y-12=68
20y=68+12
20y=80
y=80/20
y=4

5. You already have the value of "y". Now, you must substitute this value into the equation (ii):

x=8y-12
x=8(4)-12
x=32-12
x=20

6. Therefore, the result is:

x=20
y=4

3 0

3 years ago

Ellen has a bag with 3 red marbles and 2 blue marbles in it. She is going to randomly draw a marble from the bag 300 times, putt

Contact [7]

Answer:

120

Step-by-step explanation:

Blue marbles = 2

Red marbles = 3

Total marbles = 2 + 3 = 5

If a marble is drawn 300 times ;prediction for the number of times a blue marble will be drawn :

P(drawing a blue marble) = number of blue marbles / total number of marbles

= 2 /5 = 0.4

Predicted number of times blue is drawn in 300 selections :

300 * 0.4 = 120

Approximately 120 times

3 0

3 years ago

I am stuck on this question of knowing how to break it down to get the answer

Anna71 [15]

Solve and simplify:
$3 \times 3 - (( - 1) ^{2} - 2) ^{2}$
$3 \times 3 - (1 - 2) ^{2}$
$3 \times 3 - ( - 1)^{2}$
$3 \times 3 - (1)$
$3 \times 3 - 1$
$9 - 1$
$8$

Hope this helped :)

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cunderline%7B%20%5Cunderline%7B%20%5Ctext%7Bquestion%7D%7D%7D%20%3A%20" id="TexFormula1"

Inga [223]

Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

Simplify:

$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their <em>slopes will be negative reciprocals</em> of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

$x^2=2\sqrt{1-x^2}-(1-x^2)$

Distribute:

$x^2=2\sqrt{1-x^2}-1+x^2$

So:

$0=2\sqrt{1-x^2}-1$

Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

$\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}$

Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

$\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}$

And since we know that R is (0, 2), R is the y-intercept of RM. Then, using the slope-intercept form:

$y=mx+b$

We can see that the equation of Line RM is:

$y=-\sqrt{3}x+2$

6 0

3 years ago

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Simplify: (m – 3n)2

Lapatulllka [165]

2m-6n or the last one

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3 years ago

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