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3 years ago

HELP PLEASEEEE ITS DUE SOON! SHOW YOUR WORK & ILL GIVE BRAINLIEST I PROMISE

Mathematics

1 answer:

laiz [17]3 years ago

3 0

Answer:

(2,10) or x=2 y=10

Step-by-step explanation:

1. Pick one of your equations and solve for a variable. I chose the first equation and solved for x.

5x-2y=-10 (Move the -2y to the other side, you need to do the opposite so you add +2y to -10)

5x=2y-10 (Divide the 5 from the x)

x=2/5y-2

2. Now take what you got for x and plug it into the x variable on the other equation.

3(2/5y-2)+6y=66 (Multiply 3 by 2/5y and -2)

6/5y-6=6y=66 (Move the -6 to the other side and add 6/5y to 6y)

36/5y=72 (Since the number on the y is a fraction, you must do the opposite to the other side)

y=72/1 x 5/36 (Flip your fraction and multiply it by the 72)

y=10

3. Now that you have one of the variables solved for, in order to get the other we must plug in what we have to the first equation.

5x-2(10)=-10 (Multiple 2 by 10)

5x-20=-10 (Move -20 to the other side, since you do the opposite add +20 to the -10)

5x=10 ( Divide 10 by 5)

x= 2

4. If needed, plug in the values of x and y to check your solution.

Hope this could help! :)

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alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

PLZ HELP ASAP TANGENTS OF CIRCLES PROBLEMS

quester [9]

A circumscribed angle is that which is formed by the intersection of the two tangent lines in a circle. With this, we can conclude that segments AC and AB are tangent to circle O. The tangent lines forms a right angle with the radius of the circle drawn from the center of the circle to the tangent point.

By the explanation above, we can say that angles C and B are equal to 90° and that triangle ACO and triangle ABO are congruent. Which means that segment AC is equal to segment AB. Thus, the length of AB is also 4.

Answer: 4 units

7 0

4 years ago