If triangle ABC is rotated 90 counterclockwise and then reflected across the x- axis, what will be the coordinate of A?

Mathematics

1 answer:

Furkat [3]3 years ago

8 0

Answer:

Step-by-step explanation:

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Determine the center and radius of the following circle equation: x2 + y2 – 10x + 8y + 40 = 0

antiseptic1488 [7]

Answer:

The center of this circle is (5, -4) and the radius is 1.

Step-by-step explanation:

First regroup these terms according to x and y :

x^2 - 10x + y^2 + 8y = -40

Next, complete the square for x^2 - 10x: x^2 - 10x + 5^2 - 5^2.

and the same for y^2 + 8y: y^2 + 8y + 16 - 16

Substituting these results into x^2 - 10x + y^2 + 8y = -40, we get:

x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40.

Next, rewrite x^2 - 10x + 25 and y^2 + 8y + 16 as squares of binomials:

Then x^2 - 10x + 5^2 - 5^2 y^2 + 8y + 16 - 16 = -40 becomes:

(x - 5)^2 + (y + 4)^2 - 25 - 16 = -40, or:

(x - 5)^2 + (y + 4)^2 = 1

This equation has the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Matching like terms, we get h = 5, k = -4 and r = 1.

The center of this circle is (5, -4) and the radius is 1.

6 0

3 years ago

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Hpw to find exponential decay equation of radioaxtive subsrance

max2010maxim [7]

I believe you are referring to the half-life formula, which is used to calculate the half-life (when the substance drops to half of its original amount) of radioactive substances. The equation is as follows: $A = P(.5)^{t/h}$

A is the remaining amount
P is the initial amount
t is the time that has passed
h is the half-life of the substance

When plugging values into the equation, make sure that t and h are both measured in the same unit (e.g. t days and h days), as well as A and P.

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3 years ago

**MARKING BRAINLIEST PLEASE HELP**

raketka [301]

Answer:

$\text{1. } v=16t+\frac{h-c}{t},\\\text{2. }90\:\mathrm{ft/s}, \\\text{3. Cannot be determined}$

Step-by-step explanation:

1. The initial equation given to us is $h=-16t^2+vt+c$ . Rearranging the equation to isolate $v$ , we have:

$vt=16t^2+h-c,\\\boxed{v=16t+\frac{h-c}{t}}$

2. Using the equation we rearranged in part 1, we can substitute given values:

$v=16(3)+\frac{131-5}{3},\\v=48+42=\boxed{90\:\mathrm{ft/s}}$

3. We see from our equation in part 1 ( $v=16t+\frac{h-c}{t}$ ) that when $t=0$ , the denominator of our fraction will be equal to 0. Since we cannot divide by 0, the velocity remains undefined and cannot be determined.