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adell [148]
4 years ago
6

Can someone help please and thank you

Mathematics
1 answer:
denis-greek [22]4 years ago
8 0
Try making a grid like this. Apologies for the terrible drawing and handwriting. It represents all possible results of rolling two dice and summing their values. The blue column and yellow row represent the values of each die, and the green numbers reptesent their sums.
First, you can see that there are 36 possible values. Second, it's easy to see how many results are greater than 9, since they're all in one corner. And last, you can see just how many are even and how many are odd. Remember to reduce your fractions in the end.

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Suppose 255,113 people live in a city. Is it reasanoble to say that about 300,000 people live in the city? Use the number line t
motikmotik
It is reasonable. Since 255,113 is closer to 300,000 than 200,000 on the number line, you can round the number to 300,000.
3 0
3 years ago
Anyone know this i need help this is for a final
Nuetrik [128]
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5 0
3 years ago
Help!!! f(x)=x+3 g(x) =7x+4
Simora [160]
Y=0 ➡️ x= - 0.5 or x=4

5 0
3 years ago
Read 2 more answers
Please write out the problem on paper
skad [1K]
Your answer will be n5
7 0
3 years ago
Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

 magnitude of x is always positive

Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

* The solution of Ix - 3I = -12 is ∅

4 0
3 years ago
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