All categories

3 years ago

Determine if the equation 1/3×y represents a proportional relationship

Mathematics

1 answer:

Rudiy273 years ago

6 0

Answer:

Step-by-step explanation:

(1/3)y=y(1/3) is not an equation, it's just an expression.

You might be interested in

A store donated a percentage of every sale to. The total sales were $6.740 so the store donated $337. What percent of $6,740

Irina-Kira [14]

Answer - 5%

6740=100%
337= x
Criss cross so do 100x337 and then divide by 6740 and you’ll get 5 which you can check by doing 5/100 of 6740

8 0

2 years ago

A company tests two new products to make sure they last for more than a year. Product 1 had 950 out of 1,000 test items last for

zheka24 [161]

Answer:

product one; it has a higher success rate

Step-by-step explanation:

to find out which has a higher rate of success we can find out their percentage of success

we can divide their success amounts by test amounts

950 / 1000 = 95/100 = 95%

150/200 = 75/100 = 75%

product 1 has a higher success rate

5 0

2 years ago

PLEASE HELP ME ASAP !!!!!

Schach [20]

Answer:

Step-by-step explanation:

1a) the y intercept, or 20

b) from 0 weeks to 1 week the difference is 10

c) its y=mx+b m is the slope(difference) and b is the y-int so y = 10x+20

2a) the y int, is the same as the x(number of games played) being zero, so here it is 25

2b) after 3 games it was a loss of 1.50 !!! IMPORTANT !! after THREE weeks, so its 1.5/3=0.5 amount on card per week

2c) (ref 1c) but with that same formula/equation - we get y=0.5x+25

3 0

2 years ago

An investment in a savings account grows to three times the initial value after t years. If the rate of interest is 5%, compound

RideAnS [48]

3p=pe^0.05t
3=e^0.05t
T=(log(3)÷log(e))/0.05
T==21.97 years

4 0

2 years ago

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

2 years ago

Determine if the equation 1/3×y represents a proportional relationship​

Determine if the equation 1/3×y represents a proportional relationship