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3 years ago

Estimate each product.solve to check your estimate.40×6

Mathematics

2 answers:

Fed [463]3 years ago

8 0

Well 40 times 6 is 240 which means that you don't have to round.

stiv31 [10]3 years ago

4 0

All you do is cover the zero in forty and do four times six and when you get the answer add zero to the end of the number.

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Please help me with homework

Ksivusya [100]

Answer:

1.) 5/4

2.) -4/5

3.) 1/4

4.) -1

5.) 1

Step-by-step explanation:

<u>Brainliest Please</u>

7 0

3 years ago

Use the distributive property to write an expression that is equivalent to each given property.

Nadya [2.5K]

Answer:

-6x+12

-9+5a

7x+63

8y-4/5x-12

Step-by-step explanation:

To solve these all you have to do is take the number outside of the parenthesis and multiply them by each term.

-3(2x)-3(-4) All I did here was expand the problem to show what i mean by terms.

-6x+12

Do this for each problem.

0.1(-90)+0.1(50a)

-9+5a

-7(-x)-7(-9) For this one you are taking negatives times negatives so each answer will be positive.

7x+63

4/5(10y)+4/5(-x)+4/5(-15)

8y-4/5x-12

3 0

3 years ago

Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

NemiM [27]

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$

3 0

3 years ago

What is the greatest common factor of 8m, 36m", and 12?

wlad13 [49]

Answer: 4

Step-by-step explanation:

8m = 2*2*2*m

36m^3 = 2*2*3*3*m*m*m

12 = 2*2*3

GCF=2*2

Hope this helps!!

4 0

3 years ago

Given that LM and LN are tangent to the circle and that the measure of angle MLN = 75.86, find the measure of arc MN.

siniylev [52]

we know that

The measure of the external angle is the semidifference of the arcs that it covers

∠MLN=(1/2)*(mayor arc MN-minor arc MN)

Let

x------> minor arc MN

major arc MN=360-x

substitute in the formula above

∠MLN=(1/2)*(mayor arc MN-minor arc MN)

75.86=(1/2)*(360-x-x)------> 151.72=(360-2x)------>360-151.72=2x

x=104.14°

therefore

the answer is

the measure of arc MN (minor arc) is 104.14°

7 0

3 years ago