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bagirrra123 [75]
3 years ago
15

Which of the following shows the solution of 7x – 5 > 16?

Mathematics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

x<3

Step-by-step explanation:

add 5 to 16. then divide 7 from 21. the sign flips with inequalities involving multiplication and/or division

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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
Which is greater 3 miles or 5,000 yards? How much greater?
Brilliant_brown [7]
5,000 yards is greater
3 0
3 years ago
Read 2 more answers
A bag contains 2 oatmeal cookies, 1
Alex Ar [27]

Answer:

1. 33.33%

2. 16.67%

Step-by-step explanation:

there's 6 cookies in total and 2 that are oatmeal so at first the ratio is 2:6 or 33.33% then you eat one so its now 1:6 or 16.67%

7 0
4 years ago
Read 2 more answers
What is the inverse of the function f(x)= 2x+1​
Sergeu [11.5K]

In order to invert a function, switch y and x in the definition, and solve for y again:

y=2x+1 \mapsto x=2y+1

Solving for y, we have

x=2y+1\iff x-1=2y \iff y=\dfrac{x-1}{2}

8 0
3 years ago
Read 2 more answers
Please help with this problem.
larisa86 [58]

Answer:

60°, 120°

Step-by-step explanation:

\frac{ {tan}^{2}x }{2}  - 2 {cos}^{2}x = 1 \\   \\  \frac{ {tan}^{2}x  - 4{cos}^{2}x }{2} = 1 \\  \\ {tan}^{2}x  - 4{cos}^{2}x = 2 \\  \\  \frac{{sin}^{2}x}{{cos}^{2}x} - 4{cos}^{2}x = 2 \\  \\ \frac{{sin}^{2}x - 4{cos}^{4}x}{{cos}^{2}x}  = 2 \\  \\ {sin}^{2}x - 4{cos}^{4}x = 2{cos}^{2}x \\  \\ 4{cos}^{4}x  + 2{cos}^{2}x - {sin}^{2}x = 0  \\  \\ 4{cos}^{4}x  + 2{cos}^{2}x  +  {cos}^{2}x  - 1= 0  \\  \\ 4{cos}^{4}x  + 3{cos}^{2}x   - 1= 0  \\  \\ 4{cos}^{4}x  + 4{cos}^{2}x -   {cos}^{2}x - 1= 0  \\  \\4{cos}^{2}x({cos}^{2}x + 1) - 1({cos}^{2}x + 1) = 0 \\  \\ ({cos}^{2}x + 1)(4{cos}^{2}x - 1) = 0 \\  \\ ({cos}^{2}x + 1) = 0 \: or \: (4{cos}^{2}x - 1) = 0 \\  \\ {cos}^{2}x =  - 1 \: or \: 4{cos}^{2}x = 1 \\  \\ {cos}x = \sqrt{ - 1}  \: which \: is \: not \: possible \\  \therefore \: {cos}^{2}x =  \frac{1}{4}  \\  \\ \therefore \: {cos}x =   \pm\frac{1}{2} \\  \\ \therefore \: {cos}x =   \frac{1}{2}  \: or \: {cos}x =    - \frac{1}{2}  \\  \\ \therefore \: {cos}x =   {cos}60 \degree \: or \: {cos}x =     {cos}120 \degree \\  \\ \therefore \:x = 60 \degree \:  \: or  \: \: x  = 120 \degree

6 0
3 years ago
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