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iren [92.7K]
4 years ago
11

Geometry question, any help is appreciated!

Mathematics
1 answer:
Free_Kalibri [48]4 years ago
8 0
(x+2)² +(y+10)² =4
(x-h)² +(y-k)² =r²
(x-(-2))² +(y-(-10))² =2²
h=-2,
k=-10
Center is (-2, -10).
Radius r=2.

Answer C. Center is (-2, -10), r=2.
You might be interested in
Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

3 0
3 years ago
Is sqrt(25a^2/4) the same as sqrt(25a^2)/sqrt(4)?
Evgesh-ka [11]
LHS\\ \\ =\sqrt { \frac { 25{ a }^{ 2 } }{ 4 }  } \\ \\ ={ \left( \frac { 25{ a }^{ 2 } }{ 4 }  \right)  }^{ \frac { 1 }{ 2 }  }

\\ \\ =\frac { { \left( 25{ a }^{ 2 } \right)  }^{ \frac { 1 }{ 2 }  } }{ { 4 }^{ \frac { 1 }{ 2 }  } } \\ \\ =\frac { \sqrt { 25{ a }^{ 2 } }  }{ \sqrt { 4 }  } \\ \\ =RHS
6 0
3 years ago
Read 2 more answers
In ΔBCD, the measure of ∠D=90°, the measure of ∠C=77°, and CD = 41 feet. Find the length of BC to the nearest foot.
OLga [1]

Answer:

182 ft

Step-by-step explanation:

When you are not given a diagram, always draw one for yourself (see diagram).

When you have a right triangle (triangle with 90° angle), you can use the trigonometry ratios, You can remember them using SohCahToa. It is read like this:

<u>s</u>inθ = <u>o</u>pposite/<u>h</u>ypotenuse           Soh

<u>c</u>osθ = <u>a</u>djacent/<u>h</u>ypotenuse          Cah

<u>t</u>anθ = <u>o</u>pposite/<u>a</u>djacent                Toa

"θ" means the angle of reference (the angle you are talking about).

We are looking for the length of BC, which is the <u>hypotenuse</u>. I labelled it "d" (lowercase D) because it is opposite to ∠D.

We know ∠C = 77°. This will be our angle of reference (replace θ).

The side we know is DC, also known as "b" (lowercase B) because it's opposite to ∠B. "b" is the <u>adjacent</u> side when θ = C because "b" is touching ∠C.

Take the general trig. formula that has <u>hypotenuse</u> and <u>adjacent</u>: (cosine ratio)

cosθ = adjacent/hypotenuse

Substitute the variables specific for this problem.

cosC = b/d

Substitute the values you know.

cos77° = (41 ft) / d

Isolate "d" to the left side

dcos77° = d*\frac{41ft}{d}     Multiply both sides by "d"

dcos77° = 41 ft                          

dcos77° / cos77° = 41 ft / cos77°        Divide both sides by cos77°

d = 41 ft / cos77°                      Input into calculator

d = 182.261874....... ft           Unrounded answer

d ≈ 182 ft                       Rounded to nearest foot (whole number)

Remember d = BC. It's often easier to use one letter for calculations.

Therefore the length of BC is about 182 feet.

8 0
3 years ago
What is the measure of angle A
stellarik [79]

Answer:

The angle of A is 80°

Step-by-step explanation:

Given that angle A lies on a straight line which is 180°. So in order to find A, you have to substract 100° from 180° :

∠A + 100° = 180°

∠A = 180° - 100°

∠A = 80°

4 0
3 years ago
A coin is tossed twice. What is the probability of getting a tail in the first toss and a tail in the second toss?
skelet666 [1.2K]

Answer:

<h2>1/4 Chances</h2><h2>25% Chances</h2><h2>0.25 Chances (out of 1)</h2>

Step-by-step explanation:

Two methods to answer the question.

Here are presented to show the advantage in using the product rule given above.

<h2>Method 1:Using the sample space</h2>

The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below

The first toss gives two possible outcomes: T or H ( in blue)

The second toss gives two possible outcomes: T or H (in red)

From the three diagrams, we can deduce the sample space S set as follows

          S={(H,H),(H,T),(T,H),(T,T)}

with n(S)=4 where n(S) is the number of elements in the set S

tree diagram in tossing a coin twice

The event E : " tossing a coin twice and getting two tails " as a set is given by

          E={(T,T)}

with n(E)=1 where n(E) is the number of elements in the set E

Use the classical probability formula to find P(E) as:

          P(E)=n(E)n(S)=14

<h2>Method 2: Use the product rule of two independent event</h2>

Event E " tossing a coin twice and getting a tail in each toss " may be considered as two events

Event A " toss a coin once and get a tail " and event B "toss the coin a second time and get a tail "

with the probabilities of each event A and B given by

          P(A)=12 and P(B)=12

Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows

        P(E)=P(A and B)=P(A∩B)=P(A)⋅P(B)=12⋅12=14

NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.

We now present more examples and questions on how the product rule of independent events is used to solve probability questions.

8 0
3 years ago
Read 2 more answers
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