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Arlecino [84]
3 years ago
9

Quinn rode her bike 5/6 of a mile. Beth rode her bike 1/3 of a mile. How much farther did Quinn ride that Beth?

Mathematics
1 answer:
garik1379 [7]3 years ago
7 0
First multiply 1/3 by 2. 1 times 2 is 2 and 3 times 2 is six. Therefore 1/3 equals 2/6. Now subtract 5/6 by 2/6, which equals 3/6, now simplify it which makes it 1/2.

Answer: Quinn rode 1/2 a mile more than Beth.
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What is the equation of a parabola if the vertex is (1, 4) and the directrix is located at y = 7?
Oksana_A [137]

According to the vertex and the directrix of the given parabola, the equation is:

y = \frac{3}{4}(x - 1)^2 + 4

<h3>What is the equation of a parabola given it’s vertex?</h3>

The equation of a quadratic function, of vertex (h,k), is given by:

y = a(x - h)^2 + k

In which a is the leading coefficient.

The directrix is at y = k + 4a.

In this problem, the vertex is (1,4), hence:

h = 1, k = 4

The directrix is at y = 7, hence:

4 + 4a = 7

a = \frac{3}{4}

Hence, the equation is:

y = a(x - h)^2 + k

y = \frac{3}{4}(x - 1)^2 + 4

More can be learned about the equation of a parabola at brainly.com/question/26144898

4 0
2 years ago
Which is the solution of the equation a + 2 = 10?
Alexandra [31]

Hey there!

a + 2 = 10

SUBTRACT 2 to BOTH SIDES

a + 2 - 2 = 10 - 2

CANCEL out: 2 - 2 because that gives you 0

KEEP: 10 - 2 because that helps solve for the a-value

10 - 2 = a

10 - 2 = 8

Answer: Option C. a = 8

Good luck on your assignment and enjoy your day!

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8 0
3 years ago
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Label the terms in the expression as constant, linear, and quadratic
Masteriza [31]

Answer:

Step-by-step explanation:

4x^2 is the quadratic

3x is linear

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8 0
3 years ago
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.

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3 years ago
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How do i find the slant asymptope of x^3/(2(x+2)(x-4) aka x^3/(2x^2-8)
hram777 [196]
I’m really sorry I just don’t know I guess I’m. It smart enough try to look it up and wait for someone else to answer sorry again :(
6 0
2 years ago
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