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Anika [276]
3 years ago
15

What is the measures of angle ECD? Please show work.

Mathematics
2 answers:
bazaltina [42]3 years ago
6 0

The anser is 40

the resense is verticle angle are congrounent.

madreJ [45]3 years ago
4 0
The answer is D. 80°. Angle B is 60 because the angle is adjacent. That makes 60 the vertical angle for the second angle in the triangle with the 40. Since we know two of the angles, we add them together to get 100. 180-100 is 80°
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Consider this function.
charle [14.2K]

Answer:

3. Since the range of the original function is limited to y> 6, the domain of the inverse function is x ≥ 6.

Step-by-step explanation:

The domain of a function is the range of its inverse, and vice versa. The only answer choice that expresses this relationship is choice 3.

__

Comment on the answer choice:

The slope of the function is undefined at x=4, so restricting the function domain to the portion with positive slope means the domain restriction of the function is x > 4. That also means the range restriction of the function is y > 6. The domain restriction of the inverse function is the same: x > 6, not x ≥ 6. The answer choice has an error.

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3 years ago
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Is it possible for a triangle to have sides with the given lengths? Explain. 5 in., 8 in., 15 in.
Nitella [24]
<span>Triangle Inequality Theorem

</span><span>the sum of two side lengths of a triangle is always greater than the third side.
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5+8=13 
your answer = no
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3 years ago
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What is the total number of funny numbers between 30 and 60
aliina [53]

Answer:

Not exactly sure what "funny numbers" are but I'm assuming you mean prime numbers? In that case there are a total of 7 prime numbers between 30 and 60.

Step-by-step explanation:

4 0
3 years ago
Will mark brainliest!!!plz helppp
muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If f(x)=x^2-6x+3, then f(x-2)=(x-2)^2-6(x-2)+3.

Let's simplify that.

Distribute with -6(x-2):

f(x-2)=(x-2)^2-6x+12+3

Combine the end like terms 12+3:

f(x-2)=(x-2)^2-6x+15

Use (x-b)^2=x^2-2bx+b^2 identity for (x-2)^2:

f(x-2)=x^2-4x+4-6x+15

Combine like terms -4x-6x and 4+15:

f(x-2)=x^2-10x+19

We are given g(x)=f(x-2).

So we have that g(x)=x^2-10x+19.

The vertex happens at x=\frac{-b}{2a}.

Compare x^2-10x+19 to ax^2+bx+c to determine a,b,\text{ and } c.

a=1

b=-10

c=19

Let's plug it in.

\frac{-b}{2a}

\frac{-(-10)}{2(1)}

\frac{10}{2}

5

So the x- coordinate is 5.

Let's find the corresponding y- coordinate by evaluating our expression named g at x=5:

5^2-10(5)+19

25-50+19

-25+19

-6

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is a(x-h)^2+k where the vertex is (h,k).

Let's put f into this form.

We are given f(x)=x^2-6x+3.

We will need to complete the square.

I like to use the identity x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2.

So If you add something in, you will have to take it out (and vice versa).

x^2-6x+3

x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2

(x+\frac{-6}{2})^2+3-3^2

(x+-3)^2+3-9

(x-3)^2+-6

So we have in vertex form f is:

f(x)=(x-3)^2+-6.

The vertex is (3,-6).

So if we are dealing with the function g(x)=f(x-2).

This means we are going to move the vertex of f right 2 units to figure out the vertex of g which puts us at (3+2,-6)=(5,-6).

The y- coordinate was not effected here because we were only moving horizontally not up/down.

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Answer:

the answer is 12

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