!PLEASE HELP! what is the constant of proportionality in this table?

Hiroshi has 4 engines and 18 box cars. Find the ratio of engines to box cars. Write the ratio as a fraction in simplest form.

lyudmila [28]

To write a ratio as a fraction, we simply take the first number in the ratio and make it the numerator while taking the second number in the ratio and making it the denominator.

Because we want the ratio of engines to box cars, our ratio should be:

number of engines/number of box cars

When we substitute in our respective values, we get:

4/18

To simplify this ratio, we have to find the GCF, or greatest common factor of the numerator and the denominator, which in this case is 2. To simplify, we divide both the numerator and the denominator by the GCF, as follows:

4/2 / 18/2

When we simplify, we get:

2/9

Therefore, your answer is 2/9.

Hope this helps!

4 0

3 years ago

-2x + 7 Coefficient(s) : Variables(s): Constant:

Serggg [28]

Answer:

Variable = x

Coefficient = 2

Constant = 7

A variable is an unknown that holds an unknown value.

A coefficient is a number that multiplies something.

A constant is a fixed number.

Hope This Helps :)

4 0

3 years ago

For #1-8, solve the equation for the indicated variable: Solve for x: 1. 3x + 2 = 8

Karo-lina-s [1.5K]

Steps to solve:

3x + 2 = 8

~Subtract 2 to both sides

3x = 6

~Divide 3 to both sides

x = 2

Best of Luck!

5 0

3 years ago

Am i correct? calculus

Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}$

8 0

3 years ago

The radius of the large sphere is times longer than the radius of the small sphere.

jeka94

Answer:

1/3

Step-by-step explanation:

You can try it I keep getting 3 so maybe that’s the answer

8 0

3 years ago

Read 2 more answers