All categories

3 years ago

(1/3+9)*(8-3) in a fraction

Mathematics

1 answer:

neonofarm [45]3 years ago

6 0

1/3+9=(9 1/3)
8-3= (5)
9*5=45 add the 1/3 back in
45 1/3 for the fraction take 45*3=135+1= 136/3
136/3 is the answer

You might be interested in

Can someone pleaseeee help and if you’re correct i’ll give brainliest

vredina [299]

Answer:

14cm (i think)

Step-by-step explanation:

8cm= one part

6cm= the other

since there is no curve just add them together

6+8=14

3 0

2 years ago

A manager finds a moderate negative correlation between the number of computer sales and the number of microwave sales in his st

eduard

The true statement is the correlation is most likely due to a lurking variable.

<h3>What is negative correlation?</h3>

Correlation is a statistical measure used to measure the relationship that exists between two variables. Negative correlation is when there is an inverse relationship between the two variables. If one variable increases, the other variable decreases.

Assume that the store is located near a school where the students live on an allowance. So, students do not have time to buy both computers and microwaves. When students buy computers they do not have enough money to buy microwaves. This explains the negative correlation.

To learn more about negative correlation, please check: brainly.com/question/27246345

#SPJ1

5 0

2 years ago

The rational number -13/-5 lies to the right of zero on the number true or false.

Lemur [1.5K]

Answer:

true

hope it helps

PLEASE MARK BRAINLIEST

4 0

3 years ago

Read 2 more answers

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.