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3 years ago

2+2(560)+5638+163936

Mathematics

2 answers:

dmitriy555 [2]3 years ago

8 0

170696 is the answer fellow user

love history [14]3 years ago

8 0

First you multiply 2 and 560, which gives you 1120

Then you add everything together: 2+1120+5638+163936

The answer should be 170696

Hope i helped:)

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Rita and Tina each make $11 an hour working as cashiers at a supermarket. Last week, Rita worked r hours while Tina worked t hou

Alexxx [7]

Answer:

11(t+r)+35

Step-by-step explanation:

they make 11 dollars an hour so you have to multiply 11 by t hour an r hours. then only one of them worked overtime so you add 35.

5 0

2 years ago

Factor 3x^2-19x+20

dusya [7]

Answer:

(3x-4)(x-5)

Step-by-step explanation:

This is in the form

ax²+bx+c.

To factor this, we find factors of a·c that sum to b; this means factors of 3(20) = 60 that sum to -19:

60 = 1(60) or -1(-60); 2(30) or -2(-30); 3(20) or -3(-20); 4(15) or -4(-15); 5(12) or -5(-12); 6(10) or -6(-10). The only of these that sum to -19 are -4 and -15. This means we will split up -19x into -4x and -15x:

3x²-4x-15x+20

Next we group the first two terms and the last two terms:

(3x²-4x)+(-15x+20)

Factor out the GCF of each group. For the first group, this is x:

x(3x-4)

For the second group, this is -5:

-5(3x-4)

The common factor for these two groups is (3x-4):

(3x-4)(x-5)

4 0

3 years ago

13. The altitude of an equilateral triangle is 6 in. long. Find the length of a side of the triangle. (Hint: what triangles are

JulijaS [17]

Yeah the answer is in that file

6 0

3 years ago

In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

2 years ago

Thirty-five sophomores, 69 juniors and 24 seniors are randomly selected from 461 sophomores, 328 juniors and 558 seniors at a ce

LekaFEV [45]

Answer:

Stratified sampling technique(A)

Step-by-step explanation:

From the question, the population of an high school from which selection was made equals 461 sophomores, 328 juniors and 558 seniors.

35 sophomores, 69 juniors and 24 seniors are randomly selected. The technique used in selecting is Stratified sampling technique. This is because stratified sampling involves dividing the entire population into stratas and then selects a final sample randomly from the different strata. This means that a smaller part of the entire population is used as a sample in drawing conclusions for the entire population.

7 0

2 years ago