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2 years ago

Segment ET is a median of triangle FGE. Find the length of segment GT if GF=12.

Mathematics

1 answer:

yarga [219]2 years ago

5 0

Answer: The length of segment GT is 6

Step-by-step explanation: Have a nice day!✌️

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Please help! i need this asap

Westkost [7]

Philip equation
ordered data: 3o(oranges), 3g(green), 6y, 8r, 13b(black, 15b(blue
mean: 8 mode: 3 range: 12

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2 years ago

Please help!!! Select the quadratic equation that has no real solution.

ad-work [718]

Answer:

2x² + 6x + 20

Step-by-step explanation:

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1 year ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago

Without further study, you forget things as time passes. A model of human memory gives the percentage, p, of acquired knowledge

PIT_PIT [208]

Answer:

42.69% of information is acquired by the student in two weeks.

Step-by-step explanation:

We are given the following in the question:

$p = (100 -a)e^{-bt} + a$

where a and b are constants.

$a = 18\\b = 0.6$

p tells us about the percentage of acquired knowledge that a person retains after t weeks.

We have to find the percentage of knowledge acquired in two week.

We put t = 2

$p = (100 -18)e^{-(0.6)(2)} + 18\\p = 42.69$

Thus, 42.69% of information is acquired by the student in two weeks.

5 0

3 years ago

#17-1: Simplify this complex fraction. 1/3 / 6/9

Gnesinka [82]

$\huge{\boxed{\frac{1}{2}}}$

To divide by a fraction, you multiply by its reciprocal, which you find by flipping the numerator and the denominator. $\frac{1}{3} \div \frac{6}{9} = \frac{1}{3} * \frac{9}{6}$

Multiply the numerators and the denominators separately. $\frac{1}{3} * \frac{9}{6} = \frac{1*9}{3*6} = \frac{9}{18}$

Simplify by dividing both sides of the equation by $9$ . $\frac{9 \div 9}{18 \div 9} = \boxed{\frac{1}{2}}$

3 0

3 years ago

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