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Mazyrski [523]
3 years ago
9

Select the action you would use to solve 3x = 12. Then select the property

Mathematics
1 answer:
Nataly [62]3 years ago
3 0

x is 4 because 3 x 4 = 12

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Car A travels 20 mph faster than Car B. In the same time that Car A travels 208 mi, Car B travels 156 mi. Find their speeds. The
stellarik [79]

Step-by-step explanation:

Assuming the speed of car B to be x and the speed of car A to be x+20

<u>1</u><u>5</u><u>6</u> = <u>2</u><u>0</u><u>8</u>

x. x+20

156(x+20)= 208x

156x+3120=208x

3120=208x-156x

3120=52x

x=60

The speed of car A is (x)= 60

The speed of car B is(x+20)= 80

8 0
3 years ago
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They all decide to meet at school, and they each bring their two ingredients.
miskamm [114]

Step-by-step explanation:

they have 13 coups of flour 1000 grams of butter 7 cups of milk 7 cups of shugar and 6 eggs am not sure but maybe more than 4

4 0
3 years ago
Solve the equation below.<br> k - 6.2 = 24.8
Korolek [52]

Answer:

k = 31

Step-by-step explanation:

k - 6.2 = 24.8

   +6.2    +6.2 (Add 6.2 to both sides to isolate the variable: k)

__________

k = 31

4 0
3 years ago
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What type of number is V13?
Art [367]

Answer:

D. Irrational

Step-by-step explanation:

\sqrt{13}  \: is \: an \:   \purple{\bold {irrational}} \: number.

7 0
3 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
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