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diamong [38]
3 years ago
13

Solve (2x+3) ( x-7)=0​

Mathematics
2 answers:
Misha Larkins [42]3 years ago
6 0
(2x+3) (x-7)=0
Answer: X = - 3/2, 7
(this is if your solving for x I’m not sure if that’s what your asking) hope this helps!
Rus_ich [418]3 years ago
5 0

Answer:

Step-by-step explanation:

Either,

2x+3=0

2x=-3

x=-3/2

Or,

( x-7)=0​

x=7

If you found my answer useful then please mark me brainliest.

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In May, Jim’s lunch account has a balance of $58.19. If lunch costs $2.74 per day, how many days will Jim be able to buy lunch b
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4 0
3 years ago
Read 2 more answers
In the equation 0.75s-5/8 = 44, how do you combine the like terms?
garri49 [273]

Answer:

Step-by-step explanation:

In the given equation, the "like terms" are the constants 5/8 and 44.

It simplifies the math if we eliminate the fractions first.  Note that 0.75 = 6/8, so now we have:

8(6/8)s - 8(5/8) = 44).

Multiplying all three terms by 8 (above) yields

8(6s) - 8(5) = 8(44), or

48s              = 8(44 + 5), or 48s = 8(49)

Dividing both sides by 48 yields s:   s = 8(49/48)

Review "like terms:"  These are terms that have at least one characteristic in common.  5/8 and 44 are like terms because they are only constants (no variables are present).  We must add 5/8 and 44.   0.75s does not have a "like term" in the given equation.

6 0
3 years ago
Mike has $72 and he needs to buy some apples. Which of these equations would represent mike's situation of the apples cost $0.60
Alexxandr [17]

Answer:120 apples

Step-by-step explanation:

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Determine if the two triangles are congruent if so is it SSS,ASA,SAS,AAS
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How many words can be formed by using the W,X,Y,Z if repetitions is not allowed?
Setler [38]

Answer:

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Step-by-step explanation:

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We have 4 letters initially, so we can choose any 1 as our first letter. We have 4 choices for our first letter

However, once we choose our first letter, we can't use it anymore, so, for our second letter, we can only choose from the remaining 3 letters.

Furthermore, once we choose our second letter, we can only choose our 3rd letter from the remaining two letters we didn't choose yet.

Finally, our last letter will always be the one we didn't choose the last 3 times. So there is only one choice here.

Going off of this, we have four choices for the 1st letter, three choices for the 2nd letter, two choices for the 3rd letter, and one choice for the 4th letter

The way to calculate how many permutations we have without repetition is using factorials

N!

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In this case, it would be 4!

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