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1 year ago

Can someone help me solve this problem? it’s evaluating polynomials.

Mathematics

2 answers:

blagie [28]1 year ago

8 0

Answer:

answer is 6

Step-by-step explanation:

sergejj [24]1 year ago

6 0

Answer: 6

Step-by-step explanation:

Plug in 2 for t so the equation is p(2) = 2^3 - 2

This simplifies to 6

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Given logbM=3.2 logbN=-1.5 and logbP=2.4 find the value of logb(m^2N/p)

artcher [175]

The formulas:
$\log_a (xy)=\log_a x + \log_a y \\
\log_a (\frac{x}{y})=\log_a x - \log_a y \\
\log_a (x^n)n \log_a x$

Expand the expression:
$\log_b (\frac{m^2n}{p})=\log_b m^2 + \log_b n - \log_b p= 2 \log_b m+\log_b n - \log_b p$

Plug the values into the expression:
$\log_b m=3.2 \\ \log_b n=-1.5 \\ \log_b p=2.4 \\ \\
2 \log_b m+\log_b n - \log_b p=2 \times 3.2-1.5-2.4=6.4-3.9=2.5 \\ \\
\boxed{\log_b (\frac{m^2n}{p})=2.5}$

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2 years ago

5/6v+10=5 what is the answer

yulyashka [42]

Answer:

x= -3/2

Step-by-step explanation:

You would multiply both sides by 6x+10 and then that would give you 5-5)6x+10) and you would divide both sides by 5. This would give you 1=6x+10, you would then try to isolate the 6x giving you -6x=10-1 or -6x=9 and you would see if the solutions are in the defined ranges so x=-5/3 is NOT in the defined range but x=-3/2 is.

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2 years ago

Nancy started the year with $425 in the bank and is saving $25 each week. Sam started the year with $705 but is spending $15 a w

julia-pushkina [17]

Answer:

Step-by-step explanation:

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2 years ago

Solve each of the following quadratic equations 3x=0.5x^2 0=5x^2-2x+6

UNO [17]

Question 1

$3x=0.5x^{2}\\\\6x=x^{2} \\ \\ x^{2}-6x=0 \\ \\ x(x-6)=0 \\ \\ \boxed{x=0, 6}$

Question 2

$0=5x^{2}-2x+6 \\ \\ x=\frac{2 \pm \sqrt{(-2)^{2}-4(5)(6)}}{2} \\ \\ x=\frac{2 \pm \sqrt{-116}}{10} \\ \\ x=\frac{2 \pm 2i\sqrt{29}}{10} \\ \\ \boxed{x=\frac{1 \pm i\sqrt{29}}{5}}$

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2 years ago

Find all numbers with the given absolute value 1)12 2)1.7 3)3/5 4)3 1/6 5)0

telo118 [61]

It would be the negative form of those bumbers and the non-negative form of those numbers. The absolute value applies to negative numbers the same. For example absolute value of -3 is 3.

5 0

2 years ago