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3 years ago

Is (3,10) a solution of the equation y=4x?

Mathematics

1 answer:

Alexxx [7]3 years ago

4 0

Answer:

Step-by-step explanation:

(3, 10)

3 = x

10 = y

Substitute into equation

10 = 4(3)

10 = 12

No it isn't a solution as y doesn't = x

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Is the following a function?

saw5 [17]

Answer:

yes

Step-by-step explanation:

For the relation to be a function, there must only be one corresponding value of y (range) for any value of x in the domain, that is a one-to- one relationship

This graph displays this feature and is a function.

5 0

3 years ago

Read 2 more answers

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

4 0

3 years ago

Read 2 more answers

The price of a gallon of gasoline increased from 2.50 a gallon to 2.75 a gallon what was the percent increase

Vera_Pavlovna [14]

The increase in price was 10%
2.5*.1=.25
2.5+.25=2.75
hope this helps

8 0

2 years ago

(-8)(-5) = _<br> What’s the answer

Setler [38]

Answer:

Step-by-step explanation:

-8*-5=40

two negatives equal a positive

4 0

3 years ago

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Eden just bought a trough in the shape of a rectangular prism for her horses. She needs to know what volume of water to add to t

notsponge [240]

The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let w represent the width, hence:

length = w + 33, height = w - 13

Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w

V(w) = w³ + 20w² - 429w

Rate of change = dV/dw = 3w² + 40w - 429

When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423

When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118

Rate = 10118 - 5423 = 4695 in³/in

The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in

Find out more on equation at: brainly.com/question/2972832

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4 0

1 year ago