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RideAnS [48]
3 years ago
9

Which equation can be used to find the value of x?​

Mathematics
1 answer:
9966 [12]3 years ago
4 0

Answer: D

Step-by-step explanation:

The perimeter is all the sides added up

2.5x + 2.5x + 10 + 2 + 10 =>

5x + 22

It is also given that the perimeter is 10.5x

meaning that both these values are equal

5x + 22 = 10.5x

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Mindy is taking MA 385 this Fall and two sections are offered. She estimates that if
Leona [35]

solution:

P(passed)=P(Dr. J and passed)+P(Dr. S and passed) =0.4*0.8+(1-0.4)*0.6=0.68

hence probability that she was enrolled in Dr. J's section given passed

=P(Dr. J and passed)/P(passed)=0.4x0.8/0.68=0.4706


3 0
3 years ago
I need help with this<br>​
Snowcat [4.5K]

Answer:

Have u tried dividing y by x?

Step-by-step explanation:

Try that and then you should be able to find out what x equals. :)

7 0
3 years ago
Read 2 more answers
Pls help me hurry I need help!!
Citrus2011 [14]

C. 9 square roots of 2

To answer this question, it's super important that you understand the ratio of sides for special triangles. This triangle in particular, a 45-45-90 triangle, has a ratio between the legs and hypotenuse of 1:1:

Since we are given the value of the hypotenuse, we know that the value of the two sides multiplied by \sqrt{2} will be 18. Knowing this, we can write out an equation:

u*\sqrt{2} = 18

u = \frac{18}{\sqrt{2} }

<u>Multiply both sides by </u>\sqrt{2}<u> in order to get rid of the root in the denominator:</u>

u = \frac{18*\sqrt{2}}{\sqrt{2} * \sqrt{2}}

u = \frac{18*\sqrt{2} }{2}

u = 9\sqrt{2}

If you'd like me to explain how I got to the answer any further, just ask!

- breezyツ

7 0
3 years ago
What is sams ratio of free to paid minutes
Scilla [17]
Well, if Pablo gets 1 free minute for every 5 paid,
then he would get 8 free minutes for every 40 paid,
Sam is getting only 25 paid minutes for every 8 free ones
5 0
3 years ago
ILL BRAINLIEST YOU IF YOU GET IT RIGHT PLEASE HELP ME
11111nata11111 [884]

Step-by-step explanation:

R' = (-2,0)

E' = (-2,2)

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3 0
3 years ago
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