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3 years ago

Solve for y.

Mathematics

1 answer:

Illusion [34]3 years ago

4 0

Answer:

y=-1/4, 2.

Step-by-step explanation:

4y^2=7y+2

4y^2-7y-2=0

factor,

(4y+1)(y-2)=0

zero property,

4y+1=0, y-2=0,

4y=0-1=-1,

y=-1/4,

y=0+2=2.

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Three landmarks of baseballachievement are Ty Cobb’s batting average of 0.420 in 1911,Ted Williams’s 0.406 in 1941, and George B

harina [27]

Answer:

Ted Williams has the highest standardized score, hence Williams is the best of the three

Cobb is the second highest and hence, the better player

George Brett is third

Step-by-step explanation:

Given the data:

Decade : 1910 __ 1940 ____ 1970

Mean, μ: 0.266 __0.267 ___ 0.261

S/dev, σ : 0.0371 _ 0.0326 __ 0.0317

To compute the standard unit for batting averages, we obtain the standardized score for each of Cobb,Williams, and Brett.

Standardized score (Zscore) : (x - μ) / σ

Cobb, x = 0.420

Ted Williams, x = 0.406

George Brett, x = 0.390

COBB:

Zscore = (0.420 - 0.266) / 0.0371 = 4.1509

Ted Williams :

Zscore = (0.406 - 0.267) / 0.0326 = 4.2638

George, Brett = (0.390 - 0.261) / 0.0317 = 4.0694

Ted Williams has the highest standardized score, hence Williams is the best of the three

Cobb is the second highest and hence, the better player

George Brett is third

6 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: