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Lyrx [107]
3 years ago
7

What is the solution to this equation?

Mathematics
2 answers:
Vinvika [58]3 years ago
8 0

x+19= 26

x+19-19= 26-19

x= 7

Check answer by using substitution method

x+19= 26

7+19= 26

26= 26

Answer is x= 7 (D.)

velikii [3]3 years ago
7 0
The answer for this question is D
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The vertex is (2, -3).
Use the formula x= -b/2a to find the x value., which is 2.Then, substitute it into the function, wherever you see an x. Follow order operations and you get y, which is -3.
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Use the equation d=z–9 to find the value of d when z=10.<br><br> d=
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Step-by-step explanation:

d = z - 9

d = 10 - 9  ----> substitute

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3 years ago
PLEASE HELP PLEASE PLEASE
NNADVOKAT [17]

Answer: The value of x =2 units.

Explanation:

Since we have given that

AD= CD

BE= EA

So, we get that D and E are the mid-points of AC and AB respectively.

So, by Mid- point theorem , which states that the line joining the two midpoints is parallel to third side and it is half in length of third side i.e.

DE=\frac{1}{2}BC

So, we put the value of DE  and BC ,

5x-3=\frac{1}{2}\times (x+12)\\\\2(5x-3)=x+12\\\\10x-6=x+12\\\\10x-x=12+6\\\\9x=18\\\\x=\frac{18}{9}=2

Hence, the value of x =2 units.


4 0
3 years ago
Margo spends .25 of her paycheck on a new dress, 1/8 of her paycheck on shoes, and $24 on a birthday present. If she spent $60 a
stellarik [79]
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3 years ago
Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre
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Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

3 0
3 years ago
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