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Assoli18 [71]
3 years ago
14

Question: 3x +8 if x=2 FAST!!!!!

Mathematics
2 answers:
frutty [35]3 years ago
6 0

Answer:

3(2)+8= 14

Step-by-step explanation:

Gwar [14]3 years ago
5 0

Answer:

14

Step-by-step explanation:

its pretty much 6+8

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4 0
4 years ago
Help !!!! what’s the solution
murzikaleks [220]
B=0,4

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6 0
3 years ago
Help!!!
Jobisdone [24]

Answer:

In the year 2019 the number of new cars purchased will reach 15,000.

Step-by-step explanation:

t = 0 corresponds to the number of new cars purchased in 1998. If that is so, we can determine t ( time ) by making our quadratic equation here equal to 15,000 - considering that we want the year the number of cars reaches this value. t here is only the number of years to reach 15,000 cars, so we would have to add that value to 1998, to see the year that the cars will reach 15,000.

The " set up " should look like the following quadratic equation -

20t² + 135t + 3050 = 15,000 - Isolate 0 on one side,

20t² + 135t - 11950 = 0 - From here on let us solve using the quadratic equation formula,

t=\frac{-135+\sqrt{135^2-4\cdot \:20\left(-11950\right)}}{2\cdot \:20}:\quad \frac{-27+\sqrt{38969}}{8},

t=\frac{-135-\sqrt{135^2-4\cdot \:20\left(-11950\right)}}{2\cdot \:20}:\quad -\frac{27+\sqrt{38969}}{8} ... now as you can see we have two solutions, but time can't be negative, and hence our solution is the first one - about 21.3 years. 1998 + 21.3 = ( About ) The year 2019. Therefore, in the year 2019 the number of new cars purchased will reach 15,000.

3 0
4 years ago
Read 2 more answers
How many different strings of length 12 containing exactly five a's can be chosen over the following alphabets? (a) The alphabet
N76 [4]

Answer:

Part (A) The required ways are 792.

Part (B) The required ways are 101376.

Step-by-step explanation:

Consider the provided information.

Part (A) The alphabet {a, b}

The length of strings is 12 that containing exactly five a's.

The number of ways are: \frac{12!}{5!7!}

After filling "a" we have now 7 places.

For 7 places we have "a" and "b" alphabet but we already select a's so now the remaining place have to fill by "b" only.

Thus, the required ways are: \frac{12!}{5!7!}\times 1=792

Part (B) The alphabet {a, b, c}

We have selected five a's now we have now 7 places.

For 7 places we have "b" and "c".

Thus, there are 2 choices for each 7 place that is 2^7

Therefore the total number of ways are: 792\times 2^7=101376

Thus, the required ways are 101376.

7 0
4 years ago
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