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2 years ago

What is the product of (3a 2)(4a2 – 2a 9)?12a3 − 2a 1812a3 6a 912a3 − 6a2 23a 1812a3 2a2 23a 18

2 answers:

5 0

To get the product of (3a + 2) and (4a² - 2a + 9), we can apply the rules of distribution over multiplication and have 3a(4a² - 2a + 9) + 2(4a² - 2a + 9) 12a³ - 6a² + 27a + 8a² - 4a + 18 Adding like terms of the same degree, we can simplify it into 12a³ + 2a² + 23a + 18 Therefore, the answer is D<span>. </span>

liq [111]2 years ago

3 0

Answer: D. 12a^3 + 2a^2 + 23a + 18

Step-by-step explanation: This answer is 100% correct for edge.nuity.

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Two boys on bicycles, 45 miles apart, began racing directly toward each other. The instant they started, a fly on the handle bar

BlackZzzverrR [31]

Complete question is: Each bicycle have speed 10 mi/h and fly flies at 21 mi/h. How far the fly flies?

Answer:

47.25 mi

Step-by-step explanation:

The relative velocity of the bicycles is 10+10 = 20 mi/h

Time in which the two bicycles meet:

$t=\frac{45}{20} = 2.25 h$

Distance covered by fly:

$d=vt\\d=21mi/h\times 2.25h \\d=47.25 mi$

3 0

2 years ago

If (x)=3x-1 and s(x)=2x+1, which expression is equivalent to r over s (6) ?

katrin [286]

Answer:

$\large\boxed{\dfrac{3(6)-1}{2(6)+1}}$

Step-by-step explanation:

$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}\\\\\text{We have}\\\\f(x)=3x-1,\ g(x)=2x+1\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{3x-1}{2x+1}\\\\\left(\dfrac{f}{g}\right)(6)\to\text{put x = 6 to the equation of the function}\ \left(\dfrac{f}{g}\right)(x):\\\\\left(\dfrac{f}{g}\right)(6)=\dfrac{3(6)-1}{2(6)+1}$

6 0

2 years ago

Find the sum of the first<br>20 odd numbers.

konstantin123 [22]

Answer:

Hi there.....

Step-by-step explanation:

This is ur answer......

1, 3, 5, 7, 9, 11, 13, 15.................39

Therefore the total sum = 400

hope it helps you,

mark me as the brainliest....

Follow me......

7 0

2 years ago

Read 2 more answers

two angles are supplementary one angle measures 75% more than twice the measure of the other what are the measurement of the two

valkas [14]

Let's call the angles x and y. We know that supplementary angles must add to 180 degrees, so x+y=180.

Furthermore, we know that x is 75% more than 2y. We can write "75% more than" numerically as 1.75, so x = 1.75(2y).

We can simplify the second equation by multiplying 1.75 and 2 together to get x = 3.5y.

Now we can solve the system of equations by substitution by plugging 3.5y in for x in x+y=180.

(3.5y) + y = 180

Combine like terms.

4.5y = 180

Divide by 4.5

y = 40

Plug y = 40 in and solve for x.

x + 40 = 180

x = 140

The answer is 40 and 140. Hope this helps :)

3 0

2 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

2 years ago