All categories

3 years ago

The product of two consecutive integers is 420

1 answer:

6 0

(a) * (a+1) = 420
a^2 + a = 420
a^2 + a - 420 = 0
(a+21)(a-20) = 0
a = -21 OR 20

The two integers can be -21 and -20, OR 20 and 21.

You might be interested in

Solving fp<br> Algebra 2<br> I do not understand how to solve this, could someone please explain?

Olenka [21]

$f(x) = \sqrt{x + 2}$
What's The Fixed Point? Well, Let's Assume X=0, this is our point.

$f(0) = \sqrt{0 + 2} \\ f(0) = \sqrt{2} \\ this \: is \: our \: fixed \: point \: \\ (0, \sqrt{2} )$
OR

$f(p) = \sqrt{p + 2} \\ \\ or \: \sqrt{x + 2} = 0 \\ ( \sqrt{x + 2} ) ^{2} = {0}^{2} \\ x + 2 = 0 \\ x = - 2$
I'm unaware of the nature of the question, so here are some different ways a fixed point is found based on the merits of the question.

4 0

3 years ago

What os the correct question and why

mr_godi [17]

Answer:

maybe the second equation i.e(10+1.2)x=24

8 0

3 years ago

Barbara is 142 cm tall. This is 2 cm less than 3 times

krok68 [10]

The answer for your question is 45 cm long

3 0

3 years ago

What is the range of the function?

levacccp [35]

Answer:

e equels MC squared

Step-by-step explanation:

7 0

3 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

4 0

3 years ago