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4 years ago

PLEASE HELP!!! Write an absolute value equation representing all numbers x whose distance from 0 is 10 units.

1 answer:

5 0

X would be the absolute value of 10 or -10 since they both are 10 units from 0 so it can be x=l10l or x=l-10l, they both equal 10

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Blake has 2 hours between when he gets home from school and when he had to eat dinner. He spends one-fourth of this time doing h

NARA [144]

Answer:

150 minutes or 2 hours and 30 minutes

Step-by-step explanation:

One-fourth of 2 hours is 30 minutes. Multiply that number by 5 and you get 150 minutes or 2 hours and 30 minutes.

8 0

3 years ago

Read 2 more answers

Find an asymptote of this conic section. 9x^2-36x-4y^2+24y-36=0

ki77a [65]

We will begin by grouping the x terms together and the y terms together so we can complete the square and see what we're looking at. $(9x^2-36x)-(4y^2+24y)-36=0$ . Now we need to move that 36 over by adding to isolate the x and y terms. $(9x^2-36x)-(4y^2+24y)=36$ . Now we need to complete the square on the x terms and the y terms. Can't do that, though, til the leading coefficients on the squared terms are 1's. Right now they are 9 and 4. Factor them out: $9(x^2-4x)-4(y^2-6y)=36$ . Now let's complete the square on the x's. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so add it into the parenthesis. BUT don't forget about the 9 hanging around out front there that refuses to be forgotten. It is a multiplier. So we are really adding in is 9*4 which is 36. Half the linear term on the y's is 3. 3 squared is 9, but again, what we are really adding in is -4*9 which is -36. Putting that altogether looks like this thus far: $9(x^2-4x+4)-4(y^2-6y+9)=36+36-36$ . The right side simplifies of course to just 36. Since we have a minus sign between those x and y terms, this is a hyperbola. The hyperbola has to be set to equal 1. So we divide by 36. At the same time we will form the perfect square binomials we created for this very purpose on the left: $\frac{(x-2)^2}{4}- \frac{(y-3)^2}{9}=1$ . Since the 9 is the bigger of the 2 values there, and it is under the y terms, our hyperbola has a horizontal transverse axis. a^2=4 so a=2; b^2=9 so b=3. Our asymptotes have the formula for the slope of $m=+/- \frac{b}{a}$ which for us is a slope of negative and positive 3/2. Using the slope and the fact that we now know the center of the hyperbola to be (2, 3), we can solve for b and rewrite the equations of the asymptotes. $3= \frac{3}{2}(2)+b$ give us a b of 0 so that equation is y = 3/2x. For the negative slope, we have $3=- \frac{3}{2}(2)+b$ which gives us a b value of 6. That equation then is y = -3/2x + 6. And there you go!

8 0

3 years ago

The client is to receive cyclophosphamide (Cytoxan) 50 mg/kg intravenously in divided doses over 5 days. The client weighs 176 p

marishachu [46]

Answer:

800 mg

Step-by-step explanation:

As the excerpt states that the client weights 176 pounds and is to receive cyclophosphamide (Cytoxan) 50 mg/kg, we have to convert 176 pounds to kg:

1 kg → 2.20 pounds

x ← 176 pounds

x= (176 pounds * 1 kg)/2.20 pounds= 80 kg

Now, we have to determine the total amount of cyclophosphamide (Cytoxan) that the client has to receive:

50 mg/kg

For 80 kg: 80 kg *50 mg/kg= 4,000 mg

As the dose is divided over 5 days:

4,000 mg / 5= 800 mg

The client will receive 800 mg of cyclophosphamide each day.

6 0

3 years ago

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

frozen [14]

Answer:

a) The horizontal asymptote of $C(t)$ is $c = 0$ .

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of $C(t)$ is the horizontal line, to which the function converges when $t$ diverges to the infinity. That is:

$c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}$ (1)

$c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)$

$c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }$

$c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }$

$c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }$

$c = \frac{0}{3+0}$

$c = 0$

The horizontal asymptote of $C(t)$ is $c = 0$ .

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

$C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)$

Now we equalize the expression to zero:

$\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0$

$1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$5-3\cdot t^{2} = 0$

$t = \sqrt{\frac{5}{3} }\,h$

$t \approx 1.291\,h$

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}$

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

$C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

If we know that $t \approx 1.291\,h$ , then the value of the second derivative is:

$C''(1.291\,h) = -0.077$

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0

3 years ago

Azul used a number cube with the numbers 1–6 in a simulation. He signed 1–4 as a favorable outcome and 5–6 as the unfavorable ou

enyata [817]

Answer: The correct choice for this problem is answer D.

In the beginning Azul is describing probability with a number cube. A number cube has 6 possibilities. He labels 4 of them as favorable, 4/6 or 2/3. He labels 2 of them unfavorable, 2/6 or 1/3.

Choice D is the only choice that has a matching probability of 2/3.

7 0

3 years ago

PLEASE HELP!!! Write an absolute value equation representing all numbers x whose distance from 0 is 10 units.​

PLEASE HELP!!! Write an absolute value equation representing all numbers x whose distance from 0 is 10 units.