Mathematics: Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ t=18±1−4d√8 t=12±41−4d√2

Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ - id1624124148141291972, princessammarah4731

Ответ:

alexandraschwartz21

14.03.2022

Mathematics

C. $t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$

Step-by-step explanation:

We have been given an equation d=-16t^2+4t . We are asked to solve for t.

To solve for t we will rewrite our given equation in general form of equation (ax^2+bx+c=0) .

d+16t^2-4t=0

16t^2-4t+d=0

Since our given equation is quadratic, so we will use quadratic formula to solve for t.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Upon comparing our given equation with general form of equation we can see,

a=16

b=-4

c=d

Upon substituting our given values in quadratic formula we will get,

$t=\frac{--4\pm\sqrt{(-4)^2-4\cdot 16\cdot d}}{2\cdot 16}$

$t=\frac{4\pm\sqrt{16-64\cdot d}}{32}$

Upon factoring out 16 inside the square root we will get,

$t=\frac{4\pm\sqrt{16(1-4\cdot d)}}{32}$

$t=\frac{4\pm 4\sqrt{(1-4\cdot d)}}{32}$

$t=\frac{4}{32}\pm\frac{4\sqrt{(1-4\cdot d)}}{32}$

$t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$

Therefore, the solutions of t are $t=\frac{1}{8}\pm\frac{\sqrt{(1-4\cdot d)}}{8}$ and option C is the correct choice.

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Ответ:

jc95704816

14.03.2022

Mathematics

$t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)$

Step-by-step explanation:

Rearranging your quadratic to standard form, you get ...

16t^2 -4t +d = 0

For the quadratic formula, you have ...

a = 16

b = -4

c = d

so the solution is ...

$t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(16)(d)}}{2(16)}=\dfrac{4\pm 4\sqrt{1-4d}}{32}\\\\t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)$

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Ответ:

unkown77

26.05.2023

Mathematics

V = 72π in ^3

Step-by-step explanation:

The volume of a cylinder is given by the formula

V =pi r^2 h

We know the height is 8 and the diameter is 6

We need to find the radius

d= 2r

6= 2r

6/2 = 2r/2

3 =r

We can substitute in. We must keep pi as pi to be exact,

V = pi * 3^2 *8

V = pi * 9*8

V =72 pi

V = 72π in ^3

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Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ t=18±1−4d√8 t=12±41−4d√2

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