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2 years ago

Consider functions f and g vvvv

Mathematics

1 answer:

Jlenok [28]2 years ago

3 0

Answer:

C. 77x⁷ + 78x⁶ - 27x⁵

Step-by-step explanation:

Operation of functions

f(x) g(x)

f(x) • g(x) = (11x³ - 3x²)(7x⁴ + 9x³)

Simplify or expand

(11x³ - 3x²)(7x⁴ + 9x³)

Expand the expression

11x³ × 7x⁴ + 11x³ × 9x³ - 3x² × 7x⁴ - 3x² × 9x³

Multiply

77x⁷ + 99x⁶ - 21x⁶ - 27x⁵

Combine like terms

77x⁷ + 78x⁶ - 27x⁵

Hope this helps and have a nice day

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I dont really understand this

galina1969 [7]

Answer:

1. a^-b

2.a^b/a^c

3.a^c/b^c

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3 years ago

If a certain cannon is fired from a height of 8.8 meters above the ground, at a certain angle, the height of the cannonball ab

Dennis_Churaev [7]

Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

We need to find the time takes for the cannonball to strike the ground.

Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

Multiply both sides by 10.

$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$

Ignore the negative value of t as time can't be a negative number.

Thus,

$t=\frac{305+\sqrt{110273}}{98}\approx6.50$

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

6 0

3 years ago

What is 60% of 52? (SHOW WORK)

klemol [59]

Answer:

31.2

Step-by-step explanation:

i googled it

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3 years ago

Read 2 more answers

Simplify the following ratios:

jenyasd209 [6]

A) 3:4
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c) 3.5
d) 2:3

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3 years ago

Two numbers with a product of 1, such as 3/4 and 4/3 are called reciprocals

evablogger [386]

Answer:

thanks for the points i mean really. anways hope someone else can help

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2 years ago