What is the range of the function in the graph?
(I'VE NEVER BEEN GOOD AT THESE, COULD SOMEONE HELP?)

Answer:

1. option D is correct;  [tex](x+2)^2+(y+7)^2=36[/tex]  

2. option C is correct; center: (-4, 5) and radius = 11 unit

Step-by-step explanation:

The general equation of the circle is given by:

[tex](x-h)^2+(y-k)^2=r^2[/tex]                .....[1]

where,

(h, k) is the center of the circle and r is the radius of the circle.

1.

As per the statement:

The center of a circle is at (−2, −7) and its radius is 6.

⇒(h, k) = (-2, -7) and r = 6 units

Substitute these in [1] we have;

[tex](x-(-2))^2+(y-(-7))^2=6^2[/tex]  

⇒[tex](x+2)^2+(y+7)^2=36[/tex]  

Therefore, the equation of circle is, [tex](x+2)^2+(y+7)^2=36[/tex]  

2.

As per the statement:

The equation of a circle is:

[tex](x+4)^2+(y-5)^2 = 121[/tex]

We can write this as:

[tex](x-(-4))^2+(y-5)^2 = 11^2[/tex]

On comparing with equation [1] we have;

(h, k) = (-4, 5) and r = 11 units

Therefore, the center and radius of the circle is,

center: (-4, 5) and radius = 11 unit

Answer:

a. [tex]-2[/tex]

Step-by-step explanation:

The given polynomial is

[tex]x^6+3-2x^2+4x^7-4x[/tex]

We rewrite the polynomial in standard form to obtain;

[tex]4x^7+x^6-2x^2-4x+3[/tex]

The term of degree 2 in the polynomial is [tex]-2x^2[/tex].

The coefficient of this term is the constant that is multiplying [tex]x^2[/tex] which is [tex]-2[/tex]

[tex] |\Omega|=6^2=36\\
|A|=\underbrace{2}_{\text{the sum is 3}}+\underbrace{4}_{\text{the sum is 9}}=6\\\\
P(A)=\dfrac{6}{36}=\dfrac{1}{6}\approx17\% [/tex]

Answer:First, you want to solve for the equation for this problem, would be: 

0.05N + 0.10D = 20.50 

While N = The amount of nickels, and D = The amount of dimes. 

Since N = 164, and D = 123. It would add up to 287 coins. 164 + 123 = 287. 

Now that you have the number for those two variables, solve the equation for when N = 164, and D = 123. 

0.05N + 0.10D = 20.50 

0.05(164) + 0.10(123) = 20.50 

8.2 + 12.3 = 20.50 

20.50 = 20.50 

So, there is 164 nickels, and 123 dimes for your answer. 

I hope this helps you

A quadrilateral has angle measures of 52°, 94°, and 97°.

What is the measure of the fourth angle?

Enter your answer in the box.

Answer: 117

Step-by-step explanation:

HELLO!!! ^-^

The answer is 1.

If √6 is the geometric mean between 6 and another number, then the number is 1.

(I had the same question and got it right using this answer so yeah) ^-^

Have an amazing day!!! ^-^

Historical Fact:

Back in the older days, the word awful was a word used to praise someone!  That is because aw in awful comes from the origin awe which is used when someone is in awe of you!!! For example: You are AWESOME!!!

[tex]\sqrt{a}=b\iff b^2=a\\\\a\geq0\ \wedge\ b\geq0\\\\\text{therefore your answer is}\\\\\text{(A) The equation has no solutions.}[/tex]

Answer:

0,800

Step-by-step, explanation: took test and got correct

Answer:  The correct probability is (D) [tex]P(C/A)=\dfrac{13}{17}.[/tex]

Step-by-step explanation:  We are given to select the option that gives the correct probability statements using the Venn-diagram in the figure.

From the figure, we can see that there are three events, A, B and C.

And,

[tex]n(A)=1+6+7+3=17,\\\\n(B)=1+6+4+9=20,\\\\n(C)=6+4+6+7=23,\\\\n(A\cap B)=1+6=7,\\\\n(A\cap C)=6+7=13,\\\\n(B\cap C)=6+4=10,\\\\n(A\cap B\cap C)=6.[/tex]

By the law of SET THEORY, we have

[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cup B\cup C)=17+20+23-7-13-10+6\\\\\Rightarrow n(A\cup B\cup C)=36.[/tex]

So, the total number of elements in the sample space 'S' will be

n(S) = 36 + 8 = 44.

Therefore,

[tex]P(A/B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\dfrac{n(A\cap B)}{n(S)}}{\dfrac{n(B)}{n(S)}}=\dfrac{7}{44}\times \dfrac{44}{20}=\dfrac{7}{20},\\\\\\\\P(B/A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{\dfrac{n(B\cap A)}{n(S)}}{\dfrac{n(A)}{n(S)}}=\dfrac{7}{44}\times \dfrac{44}{17}=\dfrac{7}{17},\\\\\\\\P(A/C)=\dfrac{P(A\cap C)}{P(C)}=\dfrac{\dfrac{n(A\cap C)}{n(S)}}{\dfrac{n(C)}{n(S)}}=\dfrac{13}{44}\times \dfrac{44}{23}=\dfrac{13}{23},\\\\\\\\P(C/A)=\dfrac{P(C\cap A)}{P(A)}=\dfrac{\dfrac{n(C\cap A)}{n(S)}}{\dfrac{n(A)}{n(S)}}=\dfrac{13}{44}\times \dfrac{44}{17}=\dfrac{13}{17}.[/tex]

Thus, the correct probability is [tex]P(C/A)=\dfrac{13}{17}.[/tex]

Option (D) is correct.

The probability P(A/B) is 7/20, the probability P(B/A) is 7/17, the probability P(A/C) is 13/23, and the probability is P(C|A) is 13/17 option fourth is correct.

It is defined as the diagram that shows a logical relation between sets.

The Venn diagram consists of circles to show the logical relation.

For the value of probability P(A|B):

We know that:

[tex]\rm P(A/B)=\frac{P(A\cap B)}{P(B)} \Rightarrow \frac{\frac{n(A\cap B)}{n(T)} }{\frac{n(B)}{n(T)} }[/tex]

From the Venn diagram:

n(A∩B) = 7, n(B) = 20, n(T) = 44

P(A/B) = (7/44)×44/20)

P(A/B) = 7/20

For P(B|A):

[tex]\rm P(B/A)=\frac{P(B\cap A)}{P(A)} \Rightarrow \frac{\frac{n(B\cap A)}{n(T)} }{\frac{n(A)}{n(T)} }[/tex]

The values of n(B∩A) = 7, n(A) =17

P(B/A) = (7/44)×(44/17)

P(B/A) = 7/17

For P(A|C):

[tex]\rm P(A/C)=\frac{P(A\cap C)}{P(C)} \Rightarrow \frac{\frac{n(A\cap C)}{n(T)} }{\frac{n(C)}{n(T)} }[/tex]

The value of n(A∩C) = 13, n(C) = 23

P(A/C) = (13/44)×(44/23)

P(A/C) = 13/23

For P(C|A):

[tex]\rm P(C/A)=\frac{P(C\cap A)}{P(A)} \Rightarrow \frac{\frac{n(C\cap A)}{n(T)} }{\frac{n(A)}{n(T)} }[/tex]

The values of n(C∩A) = 13, n(A) = 17

P(C|A) = (13/44)×(44/17)

P(C|A) =13/17

Thus, the probability P(A/B) is 7/20, the probability P(B/A) is 7/17, the probability P(A/C) is 13/23, and the probability is P(C|A) is 13/17 option fourth is correct.

Learn more about the Venn diagram here:

brainly.com/question/1024798

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There are 54.6 inches ribbon does Ken need.

The process of combining two or more numbers is called the Addition. The 4 main properties of addition are commutative, associative, distributive, and additive identity.

Given that;

Ken is adding a ribbon border to the edge of his kite.

Here, Two aides of the kite measure 9.5 inches, while the other two sides measure 17.8.

Hence, Total length of ribbon is,

⇒ 2 × 9.5 + 2  × 17.8

⇒ 19 + 35.6

⇒ 54.6 inches

Thus, There are 54.6 inches ribbon does Ken need.

Learn more about the addition visit:

https://brainly.com/question/25421984

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