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3 years ago

Can someone answer this question please?

Mathematics

1 answer:

Keith_Richards [23]3 years ago

6 0

Answer:b=2.5555555555...

Step-by-step explanation:9b= -23, 9b/9=b, -23/9=2.55555..., b=2.55555...

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What is the difference between 4.507 and 4.517

vova2212 [387]

The difference would be 10.

5 0

3 years ago

Read 2 more answers

A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres

topjm [15]

Let $U(t)$ denote the amount of uranium-238 in the reactor at time $t$ . As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

$\dfrac{\mathrm dU}{\mathrm dt}=-kU$

where $k>0$ is constant. Separating variables and integrating both sides gives

$\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}$

Suppose we start some amount $u$ . This means that at time $t=0$ we have $U(0)=u$ , so that

$u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}$

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if $t$ is taken to be in years) for some starting amount $u$ ,

$0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}$

The half-life is the time $t_{1/2}$ it takes for the starting amount $u$ to decay to half, $0.5u$ :

$0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}$

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0

3 years ago

If r = -\sqrt{2}, and a5 = \sqrt{2}, what is the first term of the sequence?

shepuryov [24]

Answer: C

Step-by-step explanation:

Let x represent the first term (a₁)

a₁ = x

a₂ = x(-√2) = -x√2

a₃ = -x√2(-√2) = 2x

a₄ = 2x(-√2) = -2x√2

a₅ = -2x√2(-√2) = 4x

Since it is given the a₅ = √2, then we can solve: 4x =√2

Divide both sides by 4 to get: x = $\frac{\sqrt{2}}{4}$

this can also be written as: $\frac{1}{4}\sqrt{2}$

6 0

3 years ago

m∠AOB = 6x + 5, m∠BOC = 4x - 2, m∠AOC = 8x + 21

AnnZ [28]

The sum of angle AOB and angle BOC is equal tot he angle of AOC so

6x+5 +4x-2 = 8x+21

10x+3=8x+21

Subtract three and 8x from both sides

2x=18

Divide by two

x=9

Hope this helped!

6 0

3 years ago

You are standing 6 feet away from the stage and your friend is standing 7 feet away from the stage. You are standing on a platfo

coldgirl [10]

Answer:

The distance from your eyes to the stage is 8.85 ft

Step-by-step explanation:

Your distance to the stage is 6 feet

Your eyes are 6.5 feet tall

You need to find the distance of your eyes towards the stage.

This situation can be modeled using a rectangle triangle, where the adjacent side x is the horizontal distance to the stage and the opposite side h is the vertical distance of your eyes to the stage.

Observe the attached image.

Then, we look for the distance z from your eyes to the stage. This distance is the hypotenuse of the right triangle.

So, using the Pythagorean theorem we have:

$z ^ 2 = h ^ 2 + x ^ 2\\\\z = \sqrt{h^2 +x^2}$

Where

$h = 6.5\ ft\\\\x = 6\ ft$

therefore:

$z = \sqrt{(6.5)^2 +(6)^2}\\\\z =8.85\ ft$

4 0

3 years ago