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3 years ago

An inscribed angle is an angle formed by two radii that share an endpoint

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

6 0

Check the picture below.

notice, as from the left, we grab those two angles and slap them together on the right, they form a wide angle A, and a smaller inscribed angle B.

now, the blue and red segments that came together, are both making up the inscribed angle B, and also sharing the side or endpoint, in the middle of angle A.

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Which absolute value statement describes 9 degrees below zero?

miskamm [114]

Eh.....Actually.........
There are three answers,,,,,,
|-9|=9
-|9|=-9
-|-9|=9

7 0

3 years ago

Read 2 more answers

A cyclist traveled 9 kilometers per hour faster than an in-line skater. In the time it took the cyclist to travel 45 kilometers,

coldgirl [10]

Answer:

18 kmh

Step-by-step explanation:

Given: Speed of cyclist is 9 km/h more than skater´s speed.

Distance covered at the given time by cyclist is 45 km and by skater is 30 km.

Let´s assume the speed of skater be "x"

∴ Speed of cyclist will be $(x+9) km/h$

Remember; time= $\frac{Distance}{speed}$

We know the time taken by both cyclist and skater are same.

∴ we can form an equation to find the speed of skater.

Time taken by cyclist= time taken by skater

⇒ $\frac{45}{(x+9)} = \frac{30}{x}$

Multiplying both side by $(x+9)\ and\ x$

⇒ $45x= 30\times (x+9)$

Using distributive property of multiplication

⇒ $45x= 30x+270$

Subtracting both side by 30x

⇒ $15x= 270$

Dividing both side by 15

⇒ $x= \frac{270}{15}$

∴ $x= 18\ kmh$

Hence, speed of the skater is 18 kmh

6 0

4 years ago

What additional information is needed to determine that ΔABC ≅ ΔEDC using the HL method of congruence? answers: A) ≅ B) No addit

Savatey [412]

Answer:

AC congruent to EC

Step-by-step explanation:

This problem is asking specifically to prove triangles congruent using HL.

HL is a method of proving triangles congruent. HL stands for Hypotenuse-Leg. If you can show that the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg, respectively, of another triangle, then the triangles are congruent.

Looking at triangle ABC and EDC, we see that the legs AB and ED are congruent. We already have a pair of corresponding congruent legs. Now we need a pair of hypotenuses.

You need AC congruent to EC.

7 0

3 years ago

The concentration C of certain drug in a patient's bloodstream t hours after injection is given by

frozen [14]

Answer:

a) The horizontal asymptote of $C(t)$ is $c = 0$ .

b) When t increases, both the numerator and denominator increases, but given that the grade of the polynomial of the denominator is greater than the grade of the polynomial of the numerator, then the concentration of the drug converges to zero when time diverges to the infinity. There is a monotonous decrease behavior.

c) The time at which the concentration is highest is approximately 1.291 hours after injection.

Step-by-step explanation:

a) The horizontal asymptote of $C(t)$ is the horizontal line, to which the function converges when $t$ diverges to the infinity. That is:

$c = \lim _{t\to +\infty} \frac{t}{3\cdot t^{2}+5}$ (1)

$c = \lim_{t\to +\infty}\left(\frac{t}{3\cdot t^{2}+5} \right)\cdot \left(\frac{t^{2}}{t^{2}} \right)$

$c = \lim_{t\to +\infty}\frac{\frac{t}{t^{2}} }{\frac{3\cdot t^{2}+5}{t^{2}} }$

$c = \lim_{t\to +\infty} \frac{\frac{1}{t} }{3+\frac{5}{t^{2}} }$

$c = \frac{\lim_{t\to +\infty}\frac{1}{t} }{\lim_{t\to +\infty}3+\lim_{t\to +\infty}\frac{5}{t^{2}} }$

$c = \frac{0}{3+0}$

$c = 0$

The horizontal asymptote of $C(t)$ is $c = 0$ .

c) From Calculus we understand that maximum concentration can be found by means of the First and Second Derivative Tests.

First Derivative Test

The first derivative of the function is:

$C'(t) = \frac{(3\cdot t^{2}+5)-t\cdot (6\cdot t)}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}-\frac{6\cdot t^{2}}{(3\cdot t^{2}+5)^{2}}$

$C'(t) = \frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right)$

Now we equalize the expression to zero:

$\frac{1}{3\cdot t^{2}+5}\cdot \left(1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} \right) = 0$

$1-\frac{6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$\frac{3\cdot t^{2}+5-6\cdot t^{2}}{3\cdot t^{2}+5} = 0$

$5-3\cdot t^{2} = 0$

$t = \sqrt{\frac{5}{3} }\,h$

$t \approx 1.291\,h$

The critical point occurs approximately at 1.291 hours after injection.

Second Derivative Test

The second derivative of the function is:

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}-\frac{(12\cdot t)\cdot (3\cdot t^{2}+5)^{2}-2\cdot (3\cdot t^{2}+5)\cdot (6\cdot t)\cdot (6\cdot t^{2})}{(3\cdot t^{2}+5)^{4}}$

$C''(t) = -\frac{6\cdot t}{(3\cdot t^{2}+5)^{2}}- \frac{12\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

$C''(t) = -\frac{18\cdot t}{(3\cdot t^{2}+5)^{2}}+\frac{72\cdot t^{3}}{(3\cdot t^{2}+5)^{3}}$

If we know that $t \approx 1.291\,h$ , then the value of the second derivative is:

$C''(1.291\,h) = -0.077$

Which means that the critical point is an absolute maximum.

The time at which the concentration is highest is approximately 1.291 hours after injection.

5 0

3 years ago

Complete the statements about the graph of a function.

Tom [10]

Answer:

See below

Step-by-step explanation:

When a graph is symmetric about the origin, it is an odd function (Ex: y=x³)

When a graph is symmetric about the y-axis, it is an even function (Ex: y=x²)

A function is considered even if f(x) = f(−x) and it is considered odd if f(-x)=-f(x).

3 0

3 years ago