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3 years ago

Solve for x in the equation shown below

Mathematics

1 answer:

baherus [9]3 years ago

3 0

Answer:

x = 14

Step-by-step explanation:

Step 1: First, Simplify the fraction.

$\frac{4x}{7} =8$

Step 2: Next, Multiply both sides by 7.

4x = 8 × 7

Step 3: Simplify 8 × 7 to 56.

4x = 56

Step 4: Then, Divide both sides by 4.

$x=\frac{56}{7}$

Step 5: Lastly, Simplify the fraction and you're done!

x = 14

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What is the value of the expression 1/2x(6x9x1)+15

Orlov [11]

1/2x(6x9x1)+15
1/2x(6x9)+15
1/2x54+15
27+15
42

4 0

2 years ago

The Washington family is taking a family trip to Washington and has 429.5 mi left before they get to the hotel. If they are trav

Semenov [28]

Answer:

Yes He should stop for Gas

Step-by-step explanation:

Step one:

given data

total distance left=429.5miles

we want to find the amount of gallons the car will consume for 429.5 miles

we are told that for 33 miles traveled the car will consume 1 gallon of gas

hence for 429.5 miles the car will consume x gallons

cross multiply we have

x=429.5/33

x=13.0gallons

This shows that the tank will be empty for a 429.5 miles.

6 0

2 years ago

Use the laplace transform to solve the given system of differential equations. dx dt + 3x + dy dt = 1 dx dt − x + dy dt − y = et

blagie [28]

Let $X(s)$ and $Y(s)$ denote the Laplace transforms of $x(t)$ and $y(t)$ .

Taking the Laplace transform of both sides of both equations, we have

$\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s$

$\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}$

Eliminating $Y(s)$ , we get

$\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)$

Take the inverse transform of both sides to solve for $x(t)$ .

$\boxed{x(t) = \dfrac14 (1 - e^t)}$

Solve for $Y(s)$ .

$(s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}$