Ask question

Ask question

All categories

2 years ago

11

Which of the following expressions is equivalent to −56−13? Select all that apply. A. (−56)⋅(−3) B. (−65)⋅(−3) C. (−65)⋅(−13) D.

(56)÷(13) E. 56⋅(−3)
please answer soon, in the middle of a test, thx

1 answer:

Elden [556K]2 years ago

7 0

Answer:

A, D, and E.

Step-by-step explanation:

Sorry if wrong but im pretty sure. you flip the second fraction so it becomes a three. the divide bacomes a multiply so that should be correct

You might be interested in

Solve for B in the picture below

Oxana [17]

Answer:

b= -48

Step-by-step explanation:

by transposing,

43=59+b/3

b/3+59=43

b/3=43-59

b/3= -16

b= -16×3

b= -48

therefore the answer is b= -48

hope it helps!!!

PLEASE MARK BRAINLIEST..

5 0

1 year ago

Read 2 more answers

Someone plzzz help me i suck at math!!!<br>

Kay [80]

Answer:

B. $y=-sin\ x-2$

Step-by-step explanation:

We all suck at math... just lmk if you need more help...

5 0

3 years ago

14. Please help. What is the length of PQ⎯⎯⎯⎯⎯⎯⎯⎯?<br> Enter the correct value.

Pepsi [2]

Answer:

The correct value is 5.

5 0

3 years ago

On a coordinate plane, a graph has number of hours on the x-axis and number of haircuts on the y-axis. A line goes through point

Leni [432]

Answer:

Well the stylist can give one cut every 2hours or y=1/2x

Step-by-step explanation:

All you have to do is find the slope of the equation(rise/run) which ends up being 1/2 and then just turn it into an equation

6 0

2 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago