Ask question

Ask question

All categories

9 days ago

6

What divisor is represented by the synthetic division below?

You might be interested in

Natasha is planning a school celebration and wants to have live music and food for everyone who attends. She has found a band th

Svet_ta [12734]

Take 2.25 and multiply by each answer, then use 750 and record the total. Next, multiply 3.25 with each answer in ascending order, adjusting the figures until you reach the precise amount. For example, D indicates that for 760, 2.25 times 760 yields 2460, and multiplying by 3.25 gives you 2570. You save 10 dollars, enabling you to cover the costs for food and the band.

7 0

1 month ago

Read 2 more answers

What is the solution to the system of equations graphed below? Please Help (:

Zina [12379]

C. (1,5) The solution represents where two lines intersect.

5 0

1 month ago

A quadrilateral has vertices A(3,5), B(2,0), C(7,0) and D(8,5). Which statement about the quadrilateral is true?

babunello [11817]

ABCD forms a rhombus with adjacent sides that are not perpendicular.

3 0

2 months ago

Read 2 more answers

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [12381]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

7 0

3 months ago

Read 2 more answers