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2 years ago

What is (7(123456.789)4)0 simplified??

1 answer:

5 0

The answer is: " 1 " .
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Edit—Note: Since any value, raised to the "zero" power, equals "1"
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If f(x) = -x^2+3x, find (-5)

love history [14]

The A/= it means the answer is okay.

4 0

3 years ago

Miss Johnson gave a $1.50 tip to a waitress for serving a meal costing $10.00. What percent of the bill was her tip? PLZ ANSWER

nataly862011 [7]

Answer:

it's 15 percent

Step-by-step explanation:

1% = 0.10

2% = 0.20 and so on

so we know that 1% is 0.10 so if you multipy 15 x 10= 150 and if you do 1.50 divieded by 10 you would get 15 :)

(sorry if this is wrong)

3 0

3 years ago

John buys a computer with a 20% discount. If he paid $760 for the computer, how much was the discount?

Mumz [18]

If there was 20% taken off you add 20% back on and that is how you will get your answer. So take 760 plus the 20% which gives you 912 dollars

8 0

3 years ago

Read 2 more answers

Y-8= - * - 4 y=-* 4 (3,-5) no solution an infinite number of solutions

muminat

Given:

The system of equation is

$y-8=-\dfrac{1}{3}x-4$

$y=-\dfrac{1}{3}x-4$

To find:

The solution of given system of equations.

Solution:

The slope intercept form of a line is

$y=mx+b$

Where, m is slope and b is y-intercept.

Write the given equation in slope intercept form.

The first equation is

$y-8=-\dfrac{1}{3}x-4$

$y=-\dfrac{1}{3}x-4+8$

$y=-\dfrac{1}{3}x+4$ ...(i)

Here, slope is $-\dfrac{1}{3}$ and y-intercept is 4.

The second equation is

$y=-\dfrac{1}{3}x-4$ ...(i)

Here, slope is $-\dfrac{1}{3}$ and y-intercept is -4.

Since the slopes of both lines are same but the y-intercepts are different, therefore the given equations represent parallel lines.

Parallel lines never intersect each other. So, the given system of equation has no solution.

Hence, the correct option is B.

3 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago