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2 years ago

Find the sum of (6d+5)-(2-3d)

Mathematics

1 answer:

devlian [24]2 years ago

7 0

I think you have to combine like terms to solve this one. (6d - 3d) + (5 - 2) = 3d + 3

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if the morning temperature started at -7 celsius but warmed during the day to 24 celsius . What is the temperature change

Murrr4er [49]

Answer:

31° change

Step-by-step explanation:

If we want to find the change between two numbers, we need to imagine it like a number line.

<-------------0------------->

Let's plot -7 and 24 on this number line.

<---------- $-7$ --0------------24>

If we want to get from -7 to 0, we increase by 7. To get from 0 to 24, we increase by 24.

So the total change is $7 + 24 = 31$ .

Hope this helped!

8 0

3 years ago

Which might be a better way to evaluate 1150 divided by 46: drawing base-ten diagrams or using the partial quotients method ? Ex

hammer [34]

Answer:

idk sounds like a good question

Step-by-step explanation:

8 0

3 years ago

Please help answer all of the following multi-step equations

Triss [41]

Answer: first one x <-2

second one x>1

third one Use the given functions to set up and simplify

(1-4x)-x.32+4x<-3=(1-4x)-x= -5x+1

fourth one

13/14

Step-by-step explanation:

I can't rlly explain I just do, I hope this helps

8 0

2 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$