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rjkz [21]
2 years ago
9

Seven times the sum of a number and 8 equals 4​

Mathematics
2 answers:
jek_recluse [69]2 years ago
5 0
7•x+8 = 4
7x+8 =4
7x = -4
x = -4/7
gayaneshka [121]2 years ago
3 0

Answer:

n= -52/7

Step-by-step explanation:

7(n+8)=4

(7×n)+ (7×8)=4

7n +56=4

7n +56-56=4-56

7n+0= -52

7n=-52

7n/7= -52/7

n= -52/7

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Expand and simplify (2x^2)(4x+5)-8x(x^2-2)
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The answer would be 10x^2 + 16x

good luck on whatever it is you’re doing ^^
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Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water (in
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The diver will hit the water when the height is zero, so we will want to set our height function equal to zero
0 = -4t^2 + 11t + 3
we will then factor to find out value of t that make the function equal to zero
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3 years ago
A bag of skittles contains green and purple candies in the ratio of 3 green to 4 purple. If there are 24 purple candies in the b
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6 0
3 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
3 years ago
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