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3 years ago

Multiply. (3y^5)(5y^4)

Mathematics

1 answer:

IRINA_888 [86]3 years ago

5 0

Answer:

15y^9

Step-by-step explanation:

Step 1: Multiply the coefficients

3 * 5

= 15

Step 2: Add the powers.

5 + 4

= 9

Final answer: 15y^9

Mark as brainliest please, I need a few more :D

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Which two representations are considered to be only partial representations?

DaniilM [7]

Answer:

D) The graph and the table

Step-by-step explanation:

The equation and the verbal description fully represent the situation because their representation is valid for all possible x values. On the other hand, the table and the graph only represent certain x values.

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3 years ago

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annual salary is 48,000 and she earns 4% commission on all sales they make if they want to make a total of 6,000 this month how

AlekseyPX

Subtract 48,000 from 6,000 divide it by 4=10500.

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4 years ago

Slope intercept of (1,1) and (0,7)

Sonja [21]

Answer:

$\boxed {m = -6}$

Step-by-step explanation:

Use the Slope Formula to help you determine the slope of the following two points:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

(Where $(x_{1}, y_{1})$ is the first point and $(x_{2}, y_{2})$ is the second point)

-Apply both points onto that formula:

$(x_{1}, y_{1}) = (1, 1)$

$(x_{2}, y_{2}) = (0, 7)$

$m = \frac{7 - 1}{0 - 1}$

-Solve:

$m = \frac{7 - 1}{0 - 1}$

$m = \frac{6}{-1}$

$\boxed {m = -6}$

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3 years ago

The graph represents the height y, in feet, above the ground of a golf ball x seconds after it is hit.

Pachacha [2.7K]

C I think because all of them add up to it so it would be 5seconds

7 0

3 years ago

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A ballplayer catches a ball 3.4s after throwing it vertically upward. with what speed did he throw it, and what height did it re

Daniel [21]

From the equation of motion, we know,

$s=ut+\frac{1}{2}at^{2}$

Where s= displacement

u= initial velocity

a= gravitational force

t= time

Displacement is 0 since the ball comes back to the same point from where it was thrown.

A = $-9.8m/s^{2}$ since the ball is thrown upwards.

Plug the known values into the equation.

=> $0=u\left ( 3.4 \right )+\frac{1}{2}\left ( -9.81 \right )(3.4^{2})$

Solving for u gives :

u= 16.67 m/ sec ....... equation (1)

At maximum height, final velocity i.e v is 0

Time take to reach the top = $\frac{3.4}{2} = 1.7 sec$

$v^{2}=u^{2}+2as$

=> $0=(16.67)^{2}+2(-9.81)(s)$

Solving for s we get

s= 14.16 m

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3 years ago