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3 years ago

Plz help me with this question!

Mathematics

1 answer:

mart [117]3 years ago

4 0

Answer:

Coefficients are 2, 5, -1. (Coefficients are the numbers before the x)

Constant is 3 (constant is the number without any x)

2x^5+0x^4+5x^3+0x^2-x+3

You start with the biggest and end with the smallest, if anything isnt present int he original equation (for example x^4 in this case) put it in with a zero before it.

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The smallest number is 14

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A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming tha

White raven [17]

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

$\frac{dB}{dt} = rB-k$

$\Rightarrow \frac{dB}{dt} - rB=-k$ .......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor $=e^{\int p(t) dt$

$=e^{\int (-r)dt$

$=e^{-rt}$

Multiplying the integrating factor the both sides of equation (1)

$e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}$

$\Rightarrow e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt$

Integrating both sides

$\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt$

$\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C$ [ where C arbitrary constant]

$\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}$

Initial condition B=7864 when t =0

$\therefore 7864= \frac{k}{r} - Ce^0$

$\Rightarrow C= \frac{k}{r} -7864$

Then the general solution is

$B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}$

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

$B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}$

$\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0$

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

=$1573.17

4 0

4 years ago

Part B

ziro4ka [17]

The linear equation that has a slope of -7 and crosses the x-axis at (3, 0) is:

y = -7x + 21

<h3>How to find the linear equation?</h3>

A general linear equation is:

y = a*x +b

Where a is the slope and b is the y-intercept.

The slope must be equal to the limit found in part a, and you say that it is equal to -7, so the slope is -7. And for how is written the problem, I understand that it crosses the x-axis at x = 3.

Then we will have:

y = -7*x + b

Such that, when x = 3, y = 0, then:

0 = -7*3 + b

21 = b

Then the linear equation is y = -7x + 21

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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2 years ago

An article in Medicine and Science in Sports and Exercise "Maximal Leg-Strength Training Improves Cycling Economy in Previously

Hatshy [7]

Answer:

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

$299.4\leq\mu\leq 330.6$

Step-by-step explanation:

We have to construct a 99% confidence interval for the mean.

A sample of n=7 males is taken. We know the sample mean = 315 watts and the sample standard deviation = 16 watts.

For a 99% confidence interval, the value of z is z=2.58.

We can calculate the confidence interval as:

$M-z\sigma/\sqrt{n}\leq\mu\leq M+z\sigma/\sqrt{n}\\\\315-2.58*16/\sqrt{7}\leq\mu\leq 315+2.58*16/\sqrt{7}\\\\315-15.6\leq \mu\leq 315+15.6\\\\299.4\leq\mu\leq 330.6$

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

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3 years ago