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3 years ago

Given the following formula, solve for t v= u + at

Mathematics

2 answers:

Dafna1 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Alex17521 [72]3 years ago

3 0

The answer would be :
C). t=v-u/a

I hope this helped :)

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Find the center of the circle that you can circumscribe about DABC.

Vlada [557]

The center of the circle that circumscribe about DABC is $(h,k) = (1, 5)$ .

A circle can be modelled after a expression of the form:

$A\cdot x + B\cdot y + C =-x^{2}-y^{2}$ (1)

We can determine all coefficients by knowing three <em>distinct</em> points on plane.

If we know that $(x_{1}, y_{1}) = (2, 8)$ , $(x_{2}, y_{2}) = (0, 8)$ and $(x_{3}, y_{3}) = (2, 2)$ , then the solution of the system of linear equations is:

$2\cdot A + 8\cdot B + C = -68$ (2)

$8\cdot B + C = -64$ (3)

$2\cdot A + 2\cdot B + C = -8$ (4)

$A = -2, B = -10, C = 16$

Now we proceed to <em>complete</em> squares and factor each resulting perfect square trinomial in order to determine the coordinates of the center of the circle:

$x^{2}+y^{2}-2\cdot x -10\cdot y +16 = 0$

$(x^{2}-2\cdot x +1)-1+(y^{2}-10\cdot y +25)-25 +16 = 0$

$(x-1)^{2}+(y-5)^{2}= 10$

The center of the circle that circumscribe about DABC is $(h,k) = (1, 5)$ .

To learn more on circles, we kindly invite to check this verified question: brainly.com/question/11833983

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3 years ago

The area of a square garden is 50 m². How long is the diagonal?

yan [13]

D=10 m is the answer.......

3 0

4 years ago

Read 2 more answers

Solve 2sin2(x) – 5sin(x) – 3 = 0. Let u = sin(x). Which of the following is equivalent to the given equation?

Orlov [11]

Answer:

$x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z$

Step-by-step explanation:

Consider the equation

$2\sin^2x-5\sin x-3=0$

Substitute

$u=\sin x$

into the equation:

$2u^2-5u-3=0$

This is quadratic equation, where

$D=(-5)^2-4\cdot 2\cdot (-3)=25+24=49\\ \\\sqrt{D}=7$

Hence,

$u_1=\dfrac{-(-5)-\sqrt{D}}{2\cdot 2}=\dfrac{5-7}{4}=-\dfrac{1}{2}\\ \\u_2=\dfrac{-(-5)+\sqrt{D}}{2\cdot 2}=\dfrac{5+7}{4}=3$

Now, use the substitution again:

1. When $u_1=-\dfrac{1}{2},$ then

$\sin x=-\dfrac{1}{2}\\ \\x=(-1)^k\sin^{-1}\left(-\dfrac{1}{2}\right)+\pi k,\ k\in Z\\ \\x=(-1)^k\cdot \left(-\dfrac{\pi}{6}\right)+\pi k,\ k\in Z\\ \\x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z$