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3 years ago

PLEASE HELP ASAP WILL GIVE POINTS AND BRAINLIEST! TY BESTIES! Factor using GCF 20x^2+10x please show work

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer:

10x(2x+1)

Step-by-step explanation:

20x^2+10x

2*10*x*x + 10 *x

Factor out the common terms

10x(2x+1)

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Please help me solve

Law Incorporation [45]

CL ≈ AE and HG≈ZR and ∠L≈∠E as both triangles are isosceles so CL=HL and AE=EZ
so HL≈AE≈ZR≈EZ so according to side angle side ΔCLH andΔ AZE both are congruent so side CH≈AZ

4 0

3 years ago

Find the prime factorization of each number 891

cricket20 [7]

Answer:

The prime factorization of 891 is 9^2*11

Step-by-step explanation:

Divide 891 by 9, 9, and finally, 11. Or use a factorization calculator.

6 0

3 years ago

An automobile assembly line runs for 9 3/4 hours each day. employees work 3/4 hours shift. how many shifts are there in each day

babunello [35]

Answer:

the mixed fraction 9 3/4 means there are 9 4/4 s and one 3/4all togather there are 39/4 divid this by 3/4 to find the no of shifts. 39/4 ÷(3/4)

=39/4. ×4/3=13

13shifts

4 0

3 years ago

Is .09 greater then .0003

Vlada [557]

Answer:

Yes

Step-by-step explanation:

For ease, let's multiply these by 10,000.

0.09 • 10,000 = 900.00 = 900

0.0003 • 10,000 = 3.00 = 3

900 is clearly greater than 3. Therefore, 0.09 is greater than 0.0003.

8 0

3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

3 years ago