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3 years ago

Solve this question to get 11 points! 6(x + 3) = 21 x=?

Mathematics

1 answer:

azamat3 years ago

4 0

6 (x + 3) = 21
X= 3

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How many of the numbers from the set $\{1,\ 2,\ 3,\ldots,\ 50\}$ have a perfect square factor other than one?

Novay_Z [31]

Answer:

2^2 ( 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44,48

3^2(1,2,3, 5) = 9, 18, 27, 45

5^2(1, 2) = 25, 50

7^2 = 49

19 numbers

Step-by-step explanation:

5 0

4 years ago

Please help ASAP ....

vlada-n [284]

Answer:

21 3/4

Step-by-step explanation:

3 times 7 1/4 since the scale is 1:7 1/4. Good luck!! :)

7 0

3 years ago

Read 2 more answers

What three numbers equal 918

Sever21 [200]

<span>2 × 3^3 × 17

Is Your Answer
</span>

8 0

3 years ago

Can some one help pelar

7nadin3 [17]

The answer will be: x=-2

8 0

3 years ago

For each of the finite geometric series given below, indicate the number of terms in the sum and find the sum. For the value of

kotykmax [81]

Answer:

Number of term N = 9

Value of Sum = 0.186

Step-by-step explanation:

From the given information:

Number of term N = $3 (0.5)^{5} + 3 (0.5)^{6} + 3 (0.5)^{7} + \cdots + 3 (0.5)^{13}$

Number of term N = $3 (0.5)^{5} + 3 (0.5)^{6} + 3 (0.5)^{7} +3 (0.5)^{8}+3 (0.5)^{9} +3 (0.5)^{10} +3 (0.5)^{11}+3 (0.5)^{12}+ 3 (0.5)^{13}$

Number of term N = 9

The Value of the sum can be determined by using the expression for geometric series:

$\sum \limits ^n_{k=m}ar^k =\dfrac{a(r^m-r^{n+1})}{1-r}$

here;

m = 5

n = 9

r = 0.5

Then:

$\sum \limits ^n_{k=m}ar^k =\dfrac{3(0.5^5-0.5^{9+1})}{1-0.5}$

$\sum \limits ^n_{k=m}ar^k =\dfrac{3(0.03125-0.5^{10})}{0.5}$

$\sum \limits ^n_{k=m}ar^k =\dfrac{(0.09375-9.765625*10^{-4})}{0.5}$

$\sum \limits ^n_{k=m}ar^k =0.186$

6 0

3 years ago