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3 years ago

PLEASE HELP!!! According to the synthetic division below, which of the following statements are true?

Mathematics

2 answers:

mr_godi [17]3 years ago

7 0

Answer:

Options C, D and F are correct.

Step-by-step explanation:

The factors of $5x^{2}-16x+12$ are 2 and $\frac{6}{5}$

So, option C is correct. (x-2) is a factor or x=2

Option D is also correct. 2 is the root of f(x) = $5x^{2}-16x+12$

Putting 2 in place of x we get $5(2)^{2} -16(2)+12$

5*4 -32 +12=0

Option F is correct because the remainder is zero in last line.(5x-6+0)

Burka [1]3 years ago

6 0

The answer is C, D, and F. We know it is C because x-2 you would put a positive 2 and that is what is done. We know it is D because if you look at the graph that's where it would touch. We know it is F because the last number below the line is 0.

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Solve the inequality for z-9< 15. Simplify your answer as much as possible.

Fiesta28 [93]

Answer:

z < 24

Step-by-step explanation:

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2 years ago

How do I solve 20? Thanks

Gnom [1K]

20)

using
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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

What times what equals 19

andreyandreev [35.5K]

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5 0

3 years ago

Read 2 more answers

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

3 0

3 years ago