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3 years ago

Given the polynomials p(x) and g(x), which of these are guaranteed to be polynomials, too?

Mathematics

1 answer:

yarga [219]3 years ago

3 0

P(x)9(%) in fog gve VRV hobby to see what the name of the

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The game that you want to buy costs $34.95. It is on<br> sale for 20% off. What is the sale price?

lesya692 [45]

Answer:

27.96

Step-by-step explanation:

7 0

2 years ago

The quotient of a number and negative eight is five-eighths. Find the number.

max2010maxim [7]

Answer:

-64/5

Step-by-step explanation:

Basically, the word "quotient" means the answer to a division problem. Therefore, its a division problem, so you have to work backwards with multiplication. -8 times a number is 5/8, so the other number is going to be negative.

The number is -64/5

5 0

2 years ago

Read 2 more answers

Ten pictures, each measuring 4 inches by 6 inches, are mounted on a piece of green poster board so that there is no overlap. The

alexandr1967 [171]

Answer:

100 square inches

Step-by-step explanation:

Dimension of each of the ten pictures= 4 inches by 6 inches

Area of one of the pictures=4*6=24 Square Inch.

Total Area of the 10 pictures=24*10=240 Square Inches.

Dimension of the Poster Board=20 inches by 17 inches.

Area of the Poster Board=20*17=340 Square Inches.

Area not covered by the pictures=340-240=100 Square Inches

Therefore, 100 square inches of the green background will show after the pictures are mounted.

3 0

2 years ago

Necesito que alguien resuelva este ejercicio paso a paso y me lo explique como lo desarrollo y como llego a ese resultado pero l

Kobotan [32]

Answer:

translate to English please thanks!

8 0

2 years ago

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

2 years ago