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3 years ago

If angle AEB has its vertex on the center of a circle and its endpoints on the circle, what is it?

Mathematics

1 answer:

Inessa [10]3 years ago

3 0

1. Vertex E is the center of the circle, A and B are on the circle.

2. Any angle with these properties (i. E is center, AE and AB are radii) is called a central angle.

3. Check the picture.

4. An important property of these angles is that the measure of the arc AB = m(AEB)

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The following graph shows a system of inequalities.

ANTONII [103]

Answer:

Yes, the point is a solution.

Step-by-step explanation:

A system of inequalities displayed on a graph has two shaded regions, and for this example, one is red and the other is blue. The region that contains all the solutions to the system of inequalities is where the two regions intersect to usually form a new color, which in this case is purple. We can see that the point (2,2) is located within the purple region, so therefore, yes, the point is a solution.

5 0

3 years ago

True or false: Outliers can cause data sets to be skewed.

Natali [406]

True - since the outlier is way off it will cause your average to mess up which is why we test multiple times to make sure we are more accurate with our numbers

4 0

2 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

If two equations represent the same line, what can you conclude about the solution of the equations?

IceJOKER [234]

You can conclude that the equations can be equalaterial.

8 0

3 years ago

Read 2 more answers

Which compound inequality can be used to solve the inequality 13x+2/> 7

san4es73 [151]

(3x + 2) < -7 or (3x + 2 >7

Explanation/Step by step:

We have| |3x +2| >7

We divide the absolute value inequality into two inequalities

First Inequality (Which is Positive Value)
(3x + 2) >7

Second inequality (Which is the Negative Value)
-(3x + 2) >7

Then you Multiply by- 1 both sides

~And Remember that when you multiply or divide both sides of an inequality by a negative number you must reverse the inequality Symbol.~

So which that be said it would look like this:
(3x +2) < -7

So therefore (3x + 2) < -7 or (3x + 2) >7

Hope this makes since to you!!
I hope this helps as well..

5 0

3 years ago