All categories

3 years ago

What is the value of x in the equation 6x +2y=48 when y=3

Mathematics

2 answers:

ser-zykov [4K]3 years ago

6 0

Answer:

Step-by-step explanation:

2* 3 = 6

then you do 48 - 6 = 42

then divide 42 by 6 and you get 7

hope this helps :)

iren2701 [21]3 years ago

3 0

Answer:

x=7

Step-by-step explanation:

6x +2y=48

Let y=3

6x+2(3) = 48

6x+6 = 48

Subtract 6 from each side

6x+6-6=48-6

6x= 42

Divide by 6

6x/6 = 42/6

x = 7

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Step-by-step explanation:

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2 years ago

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Answer:

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2 years ago

James has 14{,}500\text { g}14,500 g14, comma, 500, start a text, space, g, end text of sand in his sandbox. He brings home anot

mariarad [96]

Question:

James has 14,500g of sand in his sandbox. He brings home another 7,400g of sand from the beach to add to his sandbox. How many kilograms of sand does James have in his sandbox now?

Answer:

The quantity of sandbox James has is 21.9kg

Step-by-step explanation:

Given

Initial quantity of sandbox = 14,500 g

Additional quantity = 7,400 g

Required

Total quantity of sandbox in kg

Tp get the total quantity of sandbox, we simply add in the initial and additional quantity of sandbox together;

This is represented mathematically as follows;

Total = Initial Quantity + Additional Quantity

Total = 14,500 g + 7,400 g

Total = 21,900 g

But the question says the answer should be in kg

1000 g is equivalent to 1 kg

21,900 g will be equivalent to $\frac{21,900}{1,000} kg$

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2 years ago

Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t > 0.

Margarita [4]

Answer:

y = 3sin2t/2 - 3cos2t/4t + C/t

Step-by-step explanation:

The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt

Comparing the standard form with the given differential equation.

p(t) = 1/t and q(t) = 3cos(2t)

I = e^∫1/tdt

I = e^ln(t)

I = t

The general solution for first a first order DE is expressed as;

y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.

yt = ∫t(3cos2t)dt

yt = 3∫t(cos2t)dt ...... 1

Integrating ∫t(cos2t)dt using integration by part.

Let u = t, dv = cos2tdt

du/dt = 1; du = dt

v = ∫(cos2t)dt

v = sin2t/2

∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt

= tsin2t/2 - cos2t/4 ..... 2

Substituting equation 2 into 1

yt = 3(tsin2t/2 - cos2t/4) + C

Divide through by t

y = 3sin2t/2 - 3cos2t/4t + C/t

Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t

3 0

3 years ago