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4 years ago

I need help. PQRS is an isosceles trapezoid. Enter coordinates for point R.

Mathematics

1 answer:

MakcuM [25]4 years ago

4 0

Step-by-step explanation:

The trapezoid is isosceles, so PQ is the same length as RS, meaning the trapezoid is symmetrical.

Therefore, R is at (c−a, b).

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Help please ! numbers 15, 16, and 17 !

timama [110]

I believe this answer is correct

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3 years ago

Evaluate (-32)^-4/5 . Do not leave any exponents in your answer.

Alexeev081 [22]

Answer:

1/16

Step-by-step explanation:

(-32)^-4/5 =
(-2^5)^-4/5 =
(-2)^-(5*4/5) =
(-2)^ -4=
16^-1=
1/16

3 0

3 years ago

What is 992,449 rounded to the nearest hundred thousand

Mila [183]

Hundred thousand is the first 9, so we will round up to make 1 million.

4 0

3 years ago

Hi please help me find for x.

eduard

Mmm I would say answer is 96
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4 0

3 years ago

Consider the differential equation,

kondor19780726 [428]

Answer:

The two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

Step-by-step explanation:

As the given equation is

$y''+7y'-8y=0\\$

So the corresponding equation is given as

$m^2+7m-8=0$

Solving this equation yields the value of m as

$(m+8)(m-1)=0\\m=-8, m=1$

Now the equation is given as

$y(t)=C_1e^{m_1t}+C_2e^{m_2t}$

Here m1=-8, m2=1 so

$y(t)=C_1e^{-8t}+C_2e^{t}$

The derivative is given as

$y'(t)=-8C_1e^{-8t}+C_2e^{t}$

Now for the first case y(t=0)=1, y'(t=0)=0

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=1\\-8C_1+C_2=0$

Solving the equation yield

$C_1=1/9 \\C_2=8/9$

So the function is given as

$y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$

Now for the second case y(t=0)=0, y'(t=0)=1

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=0\\-8C_1+C_2=1$

Solving the equation yield

$C_1=-1/9 \\C_2=1/9$

So the function is given as

$y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

So the two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

7 0

3 years ago