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2 years ago

Help plz plz plz plz plz plz

Mathematics

2 answers:

Evgesh-ka [11]2 years ago

5 0

Answer:

102

Step-by-step explanation:

1^2+2=2

2^2+2=6

3^2+2=11

10^2+2=102

s344n2d4d5 [400]2 years ago

4 0

Answer:

brainly have anything and everything that you need just ask the question and it will give the answer that they have

Step-by-step explanation:

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Plss finish fast the question is linked thank you for ur help

kobusy [5.1K]

Answer: B. 2 2/5

Step-by-step explanation: Hope this help :D

3 0

3 years ago

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Yesterday you walked 1/3 mile .today you walked 1/2 mile. How far did you walk in those two days?

Makovka662 [10]

You walked 5/6 mile

1/3(x2)=2/6
1/2(x3)=3/6
5/6 mile

7 0

2 years ago

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I need help with question 18 PLS!!!!

dolphi86 [110]

Answer:

162

Step-by-step explanation:

( -a ) ( b ) ( -a + b )

-6 ( 3 ) ( -6 + -3 )

-6 ( 3 ) ( -9 )

- 18 ( -9 )

162

6 0

2 years ago

The image of a dust mite from a scanning electron microscope is 1.5×10 to the 2nd millimeters wide. The image is 5×10 to 2nd tim

Vinil7 [7]

600 millimeters is the accuracy

7 0

2 years ago

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Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

2 years ago