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NeTakaya
3 years ago
14

What is number 32 PLEASE answer asap

Mathematics
1 answer:
xxMikexx [17]3 years ago
4 0
6:18 to 10:30
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Can anyone help?....
stepan [7]
<span>The answer two this question is C.

You can get this answer by multiplying the two polynomials and then simplifying. See the steps below:
(4x^2 - 5)(x + 3) ---> Now multiply the two parenthesis. 
4x^3 - 5x + 12x^2 - 15 ---> Arrange the numbers by decreasing  exponents. 
4x^3 + 12x^2 - 5x - 15

No simplifying can be done since there are no like terms. </span>
8 0
3 years ago
Read 2 more answers
An equation is shown below: −2(4x − 1) − 7 = 5 Which statement shows a correct next step in solving the equation?
LuckyWell [14K]
B. The equation can become −2(4x − 1) = 12 by applying the addition property of equality because we have: <span>−2(4x − 1) − 7 = 5 / + 7 <=> -2(4x-1) = 12;</span>
6 0
4 years ago
Joan and her brother are saving their money to buy a stereo that costs $250. If Joan has $195.50 and her
Rina8888 [55]

Answer:

yes, they do

Step-by-step explanation:

added together it equals 254, so unless there is tax they have enough

3 0
3 years ago
Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

5 0
3 years ago
The value of x is: <br><br> A. 40 <br><br> B. 20 <br><br> C. 13
Lady_Fox [76]
5/8 = x/32

8 x 4 = 32

5 x 4 = 20

5/8 = 20/32


x = 20
3 0
3 years ago
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