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3 years ago

I need help with the question that is attached.

Mathematics

1 answer:

aleksley [76]3 years ago

8 0

Answer:

i think its 12x^3

Step-by-step explanation:

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HELP ME PLZ IN NEED THIS DONE BY 8:10

Stella [2.4K]

Answer 16.

Step-by-step explanation:

First you replace the xs with 1/2

(√2•1/2+3)/(6•1/2-5)

Then you solve the multiplications

2•1/2=1

6•1/2=3

Now write them in

(√1+3)/(3-5)

Simplify

(√4)(-2)

Root of 4 equals to

2/-2

And the answer is -1

Answer 17:

Replace x with 8

1/2•8^2-(1/4•8+3)

Solve parenthesis, multiplication first

1/2•8^2-(2+3)

1/2•8^2-(5)

1/2•8^2-5

Solve the 8^2 and the 1/2

1/2•64-5

32-5

The answer is 17

3 0

3 years ago

Find the area of the sector formed by central angle

Alex777 [14]

Answer: 0.2pi

Step-by-step explanation:

1. Find the area of the entire circle

2. Set up a proportion that compares the relationship of the Area of sector and the Area of circle to the Arc measure and the circle measure

3. Solve!

6 0

3 years ago

In the expression 16x + 2x – 5y + 17, which terms are “like terms”?

Len [333]

The like terms are 16x and 2x

7 0

4 years ago

Read 2 more answers

In ΔRST, r = 52 cm, s = 55 cm and ∠T=48°. Find the length of t, to the nearest centimeter.

quester [9]

Answer:

Step-by-step explanation:

delta math

6 0

3 years ago

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

3 years ago