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lions [1.4K]
3 years ago
14

Ptx-icor-stv meet buddys........​

Mathematics
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

ill join you next got it

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Given angle A B C , in which Line AD bisects angle BAC. If m angle B=64 and m angle C =42, determine m angle BAD.
Andrew [12]
D=54 i might be wrong
3 0
4 years ago
The ratio of savings to expenditure is 2:8 find the savings if the expenditure is 24,000​
WITCHER [35]

Answer:

the savings is 6000

Step-by-step explanation:

We are told that the ratio of savings to expenditure is 2: 8, that is, that person saves 2 when he spends 8.

They tell us to find the savings when the cost is 24,000, so we are left with:

24000 * 2/8 = 6000

which means that when 24000 are spent the savings is 6000

6 0
3 years ago
HELP NOW PLS what is (x+6)
crimeas [40]

x+6 is x+6. It's an expression. Not sure what you need help with.

7 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
63 decreased by twice Vidya's score Use V to represent Vidya's score.
taurus [48]
V = 126 if that's what you're looking for
6 0
3 years ago
Read 2 more answers
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