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3 years ago

Which expressions correctly show a subtraction problem rewritten as an addition problem?

Mathematics

2 answers:

LiRa [457]3 years ago

8 0

First, second and fourth equation

svlad2 [7]3 years ago

6 0

Answer:

uhh i think the 2nd and the very last one

Step-by-step explanation:

sorry if its wrong

You might be interested in

Tired of WRESTLING, need you to pend this problem to the canvas... -1/2~3/4= ? "Divide"

Alexxx [7]

-2/3 is the simplified form of the expression -1/2 ÷ 3/4.

<h3>What is the simplified form of the given expression?</h3>

Given the expression in the question;

-1/2 ÷ 3/4

To divide by a fraction, multiply by its reciprocal.

-1/2 ÷ 3/4

Reciprocal of 3/4 is 4/3

Hence

-1/2 × 4/3

Next, we cancel out the common factor of 2

Factor out 2 from 4

-1/2 × 2(2)/3

-1/1 × 2/3

( -1 × 2 ) / ( 1 × 3 )

(-2) / (3)

-2/3

Therefore, the simplified form is -2/3

Learn how to solve more fraction problems here: brainly.com/question/1627825

#SPJ1

8 0

1 year ago

Estimate how much MORE water it would take to completely FILL the vase to the top?

MariettaO [177]

Answer:

81πinc^3

Step-by-step explanation:

Volume of glass = π(3)^2(12) = 108πin^3

Volume of water in glass = π(3)^2(3) = 27πin^3

Volume needed to fill completely = 108π - 27π = 81π^3

Topic: Mensuration

If you like to venture further, please give my Instagram page a follow (learntionary). I'll be constantly posting math tips and notes! Thanks!

6 0

3 years ago

A pet supply chain called Pet City has 5 hamsters and 10 gerbils for sale at its Lanberry location. At its Milford location, the

Vlad1618 [11]

Answer:

Milford location has a higher ratio of hamsters to gerbils.

Step-by-step explanation:

Given:

A pet supply chain called Pet City has 5 hamsters and 10 gerbils for sale at its Lanberry location.

At its Milford location, there are 13 hamsters and 16 gerbils.

Now, to find the location who has a higher ratio of hamsters to gerbils.

Ratio of hamsters to gerbils at Lanberry location = 5:10.

$=\frac{5}{10}=0.50$

Ratio of hamsters to gerbils at Milford location= 13:16.

$=\frac{13}{16} =0.8125$

So, 0.8125 > 0.50.

Thus, 13:16 > 5:10.

Therefore, Milford location has a higher ratio of hamsters to gerbils.

4 0

3 years ago

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.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago

Simplify: 25x - 6X - 18x + 7

azamat

Answer:

x+7

Step-by-step explanation:

25x - 6x - 18x + 7

x + 7

6 0

3 years ago