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3 years ago

What are the next two numbers in this pattern 7, 14, 28, 56.....

Mathematics

2 answers:

Svetlanka [38]3 years ago

7 0

Your answer would be A. you are multiplying by 2 each time

gregori [183]3 years ago

5 0

Answer:

Your answer is B

Step-by-step explanation:

This pattern is going by multiples, so like 7 x 8 = 56

and 7 x 9 = 63 and so on

You might be interested in

What is 10^{2} /19^{4}

MatroZZZ [7]

Answer:

$\frac{100}{130321}$

Step-by-step explanation:

10² ÷ $19^{4}$

100 ÷ $19^{4}$

100 ÷ 130321

$\frac{100}{130321}$

4 0

3 years ago

Will mark brainly

Taya2010 [7]

To answer this, we must look at the general direction of the x and y.

As x approaches infinity, we see that y also approaches infinity. This means that the 5th option with the two sentences stating this is correct.

Hope this helps x

5 0

3 years ago

Clyde simplified ( j^-8)^-2

Lina20 [59]

ANSWER

${j}^{16}$

EXPLANATION

The given expression is

$( {j}^{ - 8} )^{ - 2}$

To simplify this expression, we use the law of exponents.

Recall that:

${(a}^{m} )^{n} = {a}^{mn}$

We need to multiply the exponents to get:

$( {j}^{ - 8} )^{ - 2} = {j}^{ - 8 \times - 2}$

$( {j}^{ - 8} )^{ - 2} = {j}^{16}$

Therefore after simplification, Clyde got:

${j}^{16}$

5 0

3 years ago

Read 2 more answers

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$