The sun of 3 less than 5 times a number and the number increased by 9 is 24. What is the number.?

All of the choices are in slope-intercept, or y=mx+b form where m is the slope and b is the y intercept. Therefore if the slope is 2, and the y intercept is 8 the equation should be y=2x+8.

3 0

3 years ago

FASTTTTTTTTT PPPPPPPPPPPPPPLLLLLLLLLLLLLLLLZZZZZZZZZZZZZZZ

Deffense [45]

Answer:

(a) B

(b) $2

Step-by-step explanation:

(a) Let's say the cost of a ticket is t and the cost of popcorn is p. Then we can write the two equations from the table:

12t + 8p = 184

9t + 6p = 138

We need to solve this, so let's use elimination. Multiply the first equation by 3 and the second equation by 4:

3 * (12t + 8p = 184)

4 * (9t + 6p = 138)

We get:

36t + 24p = 552

Subtract the second from the first:

36t + 24p = 552

- 36t + 24p = 552

________________

0 = 0

Since we get down to 0 = 0, which is always true, we know that we cannot determine the cost of each ticket because there is more than one solution (infinitely many, actually). The answer is B.

(b) Our equation from this, if we still use t and p, is:

5t + 4p = 82

Now, just choose any of the two equations from above. Let's just pick 9t + 6p = 138. Now, we have the system:

5t + 4p = 82

9t + 6p = 138

To solve, let's use elimination again. Multiply the first equation by 6 and the second one by 4:

6 * (5t + 4p = 82)

4 * (9t + 6p = 138)

We get:

30t + 24p = 492

36t + 24p = 552

Subtract the second from the first:

36t + 24p = 552

- 30t + 24p = 492

________________

6t + 0p = 60

So, t = 60/6 = $10. Plug this back into any of the equations to solve for p:

5t + 4p = 82

5 * 10 + 4p = 82

50 + 4p = 82

4p = 32

p = 32/4 = $8

So the ticket costs 10 - 8 = $2 more dollars than the popcorn.

5 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

The sun of 3 less than 5 times a number and the number increased by 9 is 24. What is the number.?​

The sun of 3 less than 5 times a number and the number increased by 9 is 24. What is the number.?