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3 years ago

-5/3 is a root of f(x) = (3x + 5) (x2 - 6x + 9)2 = 0 True False

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

8 0

False...................

Mama L [17]3 years ago

5 0

For this case we have the following function:

$f (x) = (3x + 5) (x ^ 2-6x + 9) ^ 2$

To find the roots we must equal the function to 0, that is:

$(3x + 5) (x ^ 2-6x + 9) ^ 2 = 0$

So:

$3x + 5 = 0$

Subtracting 5 from both sides:

$3x = -5$

Dividing between 3 on both sides:

$x = - \frac {5} {3}$

Thus, one of the roots is $x = - \frac {5} {3}$

Answer:

True

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How many quarters doesTed have?

Zigmanuir [339]

Answer:

well in cerious george movie he didnt have any lol

Step-by-step explanation:

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3 years ago

What is the area of this polygon? Please explain!

Anna71 [15]

to find the area, it helps to know all the lengths.

The vertical (V) piece near the 9 equals 2 and the horizontal (H) piece equals 6.

(to find the V piece, I did 10-8; for the H piece I did 15-9)

Next you have to find the area. I find its easiest to split them up into more shapes so I had a big rectangle and the little square piece at the bottom.

The measurements to multiply for the rectangle are 15*8= 120

the measurements for the little square are 9*2=18

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3 years ago

A department store offers free samples of a 2-ounce

Nataly_w [17]

Answer:

The situation you are describing can be written as y=2x, or in this case t=2f, for every fragrance purchased there are 2 ounces of lotion that are given. Since the ounces of fragrance are not shown in this question, we can only make the graph on the ounces of lotion for every fragarance purchased. The points of the graph represent that for every fragrance bought, there is twice the amount of that in ounces of lotion. In summary, the rule is t=2f where t would equal purchases of fragrance and f would equal the ounces of lotion.

(If there is any more you would like to add to the question I can answer that.)

8 0

3 years ago

Apyrotechnician plans for two fireworks to explode together at the same

olga55 [171]

First note down the relevant variables from the question.
Ua (Initial velocity a) = 320ft/s
Ub (initial velocity b) = 240ft/s
Aay (acceleration of a in the vertical axis) = Aby = -32.17ft/s/s

We want to know when they will be at the same height so should use the formula for displacement:
s = ut + 1/2 * at^2

We want to find when both firework a and firework b will be at the same height. Therefore mathematically when: say = sby (the vertical displacements of firework A and B are equal). We also know that firework B was launched 0.25s before firework A so we should either add 0.25s to the time variable for the displacement formula for firework B or subtract 0.25s for firework A.

SO:
Say = Sby
320t + 1/2*-32.17t^2 = 240(t+0.25) + 1/2 * -32.17(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t^2 + 0.5t + 6.25)
320t - 16.085t^2 = 240t + 60 -16.085t^2 - 8.0425t - 100.53
320t - 240t - 8.0425t - 16.085t^2 + 16.085t^2 = 60 - 100.53
71.958t = -40.53
t = -0.56s (negative because we set t before Firework A was launched)

Now we know both fireworks explode 0.56 seconds AFTER fireworks B launches (because we added 0.25 seconds to the t variable in the equation above for the vertical displacement of Firework B).

You could continue on to find the displacement they both explode at and verify the answer by ensuring that it is equal (because the question stated they should explode at the same height by substituting the value we found for t of 0.56s into the vertical displacement formula for firework A and t+0.25s=0.81s into the same formula for Firework B

Verification:
Say = ut + 1/2at^2
Say = 320*0.56 + 1/2*-32.17*0.56^2
Say = 179.2 + -5.04
Say = 174.16ft

Sby = ut + 1/2at^2
Sby = 240*0.81 + 1/2*-32.17*0.81^2
Sby = 194.4 - 10.5
Sby = 183.9ft

While Say is close to Sby I would have expected them to be almost perfectly equal… can you please check if this matches the answer in your textbook? There could be wires due to rounding. I also usually work in SI units which use the metric system and not the imperial system although that shouldn’t make a difference. The working out and thought process is correct though and this is why trying to verify the answer is an important step to make sure it works out.

Answer: 0.56s (I think)

3 0

2 years ago

Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approxi

Semenov [28]

Answer:

Option B - $=8.4672\times 10^{13}$

Step-by-step explanation:

Given : Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5,880,000,000,000 miles.Suppose a star is 14.4 light years from earth.

To find : In scientific notation, how many miles away a star is from earth?

Solution :

One light-year is the distance that light can travel in one year is = 5,880,000,000,000 miles.

In scientific notation, $5.88\times 10^{12}$ miles.

In 14.4 lights year the distance is $=14.4\times5.88\times 10^{12}$

In 14.4 lights year the distance is $=84.672\times 10^{12}$

or $=8.4672\times 10^{13}$

Therefore, Option B is correct.

5 0

3 years ago