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Naily [24]
3 years ago
11

Bill has $2.00 in quarters and dimes. the number of quarters is 4 less than twice the number of dimes. find the number of coins

of each type
Mathematics
1 answer:
Dafna1 [17]3 years ago
7 0
Define your variables:
Q: number of quarters
D: number of dimes

Next, create your equations
Q = 2D-4
25Q+10D=200

Because one variable (Q) is already written in terms of the other variable, D, you can already start solving with substitution.
25(2D-4)+10D=200
50D-100+10D=200
60D=300
D=5

Substitute D back into the other equation to find Q. 
Q=2D-4
Q=2(5)-4
Q=6

Substitute D and Q into the second equation to check.
25Q+10D=200
25(6)+10(5)=200
150+50=200

This works, so the answer is right! Bill has 6 quarters and 5 dimes.
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1.1331

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The answer is 6.4*10^22
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Please help WOrth 10 points
rjkz [21]

Answer:

52 cubic units

Step-by-step explanation:

1. Divide the shape into 2 parts:

The first one: a 2 by 4 by 6 rectangular prism.

The second one: a 1 by 4 by 1 rectangular prism.

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6 0
3 years ago
Pls answer this question
eimsori [14]

Answer:

1) B

Step-by-step explanation:

1)

The formula for the volume of a prism is h * l * w

So, if we do that we get 25/18

2) B

Lets say we talk the top surface, there are 30 cubes on that surface and we want to find the volume of one cube so we are going to divide by 50. we take the measurements which are 5/6 and 1 for the top surface and we multiply. We get an answer of 5/6 now we divide by 30

So...

5/6 * 1/30 (We switched the numbers around because we are dividing.)

5/6 * 1/30 = 5/180 now we can simplify this to 1/36.

5 and 180 are a factor of 5

Hope this helps

8 0
3 years ago
Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point
Maksim231197 [3]

Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

Taking square on both the sides , we get

(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0

Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0
3 years ago
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