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3 years ago

What is the product? (4y - 3) (2y^2 +3y - 5)

Mathematics

1 answer:

igomit [66]3 years ago

8 0

(4y-3)(2y²+3y-5)

First , let's start with "4y"

4y*2y² = 8y³

4y*3y = 12y²

4y*-5 = -20y

Next, let's multiply by "-3"

-3*2y² = -6y²

-3*3y = -9y

-3*-5 = 15

Now, let's combine all of our values.

8y³+12y²-6y²-20y-9y+15 = 8y³+6y²-29y+15

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Answer:

Parent Function: $y=\sqrt{x}$
Horizontal shift: right 3 units
Vertical shift: up 3 units
Reflection about the x-axis: none
Vertical strech: streched

Step-by-step explanation:

assume that $y=\sqrt{x}$ is $f(x)=\sqrt{x}$ and $y=\sqrt{-2x+6}+3$ is

$g(x)=\sqrt{-2x+6}+3\\ f(x)=\sqrt{x} \\g(x)=\sqrt{-2x+6}+3$

The transformation from the first equation to the second equation can be found by finding a,h and k for each equation.

$y=a\sqrt{x-h}+k$

factor a 1 out of the absolute value to make the coefficient of x equal to 1

$y=\sqrt{x}$

factor a 2 out of the absolute value to make the coefficient of x equal to 1

$y=\sqrt{2}\sqrt{x-3}+3$

find a, h and k for $y=\sqrt{2}\sqrt{x-3}+3$

$a=1.41421356\\ h=3\\k=3$

the horizontal shift depends on the value of h when $h > 0$ , the horizontal shift is described as:

$g(x)=f(x+h)$ - the graph is shifted to the left h units

$g(x)=f(x-h)\\$ - the graph is shifted to the right h units

the vertical shift depends on the value of k

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3 years ago

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is: