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3 years ago

I love school so much that I always get straight A's. If any of y'all need help...Im here!

Mathematics

2 answers:

Fynjy0 [20]3 years ago

4 0

Help would be greatly appreciated!

love history [14]3 years ago

3 0

You have proof? What grade are you in? Text me on snap cus if you are in 8th grade I’ll need help on my geometry pleaseeee. Snap is gomezbaby_1820

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Find the value of x

Tasya [4]

because its isosceles so 9+9=x²

and x equal

2√3

8 0

3 years ago

Read 2 more answers

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

3 years ago

Point A is the point of concurrency of the angle bisectors of ΔDEF. Point A is the point of concurrency of triangle D E F. Lines

fiasKO [112]

Answer:

ZA = 3 cm

Step-by-step explanation:

Point A is the incenter of the triangle, so segments AX, AY, and AZ are radii of the circle. They are all the same length, given as 3 cm and confirmed by 3-4-5 right triangle AYD.

ZA = 3 cm

8 0

3 years ago

For which survey is a sample not necessary?

BARSIC [14]

Answer:

<h2>How many car accidents involve airbag malfunctions?</h2><h2 />

This is because there's no need to survey an entire population for accurate results.

The other 3, regarding Colorado residents, middle school students, and classmates are considered a population.

For testing the amount of car accidents, a population is not needed.

<em />

<em>I hope this helped you!</em>

7 0

3 years ago

Read 2 more answers

Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01001. In each code word, a bit is z

AleksandrR [38]

Answer:

the probability that a code word contains exactly one zero is 0.0064 (0.64%)

Step-by-step explanation:

Since each bit is independent from the others , then the random variable X= number of 0 s in the code word follows a binomial distribution, where

p(X)= n!/((n-x)!*x!*p^x*(1-p)^(n-x)

where

n= number of independent bits=5

x= number of 0 s

p= probability that a bit is 0 = 0.8

then for x=1

p(1) = n*p*(1-p)^(n-1) = 5*0.8*0.2^4 = 0.0064 (0.64%)

therefore the probability that a code word contains exactly one zero is 0.0064 (0.64%)

3 0

3 years ago