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monitta
3 years ago
8

SEE IMAGE! Two lines below intersect to form four angles, as shown. The measure of angle 2 is 114°. What is the sum of the measu

res of angles 1 and 3?
A) 66
B) 90
C) 114
D) 132

Mathematics
1 answer:
o-na [289]3 years ago
8 0
ANSWER:


Angles 2 and 4 are vertical angles, which means they have the same angle measure because they are opposite of each other. So angles 2 and 4 are both 114 so that makes 228. All of the angles have to make 360, so 360 - 228 = 132. Then, 132/2 because there are two angles. That makes 66. And angles 1 and 3 are also vertical angles so they are also the same

The answer is 132
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Given a term in a geometric sequence and the common ratio find the first five terms, the explicit formula, and the recursive formula. Given two terms in a geometric sequence find the 8th term and the recursive formula. Determine if the sequence is geometric. If it is, find the common ratio.

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What is the minimum value of the function g(x) = x2 - 6x – 12?<br> -21<br> 3-21<br> 3<br> 3+21
balandron [24]

Answer:

- 21

Step-by-step explanation:

The minimum value occurs at the vertex of the function

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

To obtain this form use the method of completing the square

Given

f(x) = x² - 6x - 12

add/ subtract ( half the coefficient of the x- term)² to x² - 6x

f(x) = x² + 2(- 3)x + 9 - 9 - 12

     = (x - 3)² - 21

with vertex = (3, - 21 )

The minimum is the value of k, that is minimum value = - 21

4 0
3 years ago
. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,
Eva8 [605]

Answer:

Elements are of the form

 (i) [a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}

(ii) [a,b)=\{[x,y) :a\leq x

(iii)(a,a]=\{(x,y] :a

(iv)(a,a)=\{(x,y): a

(v) (a,b) where a>b=\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}

(vi) [a,b] where a>b=\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi)  [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........

Because there is no increment so if a\in \mathbb R then the set  (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

(a,b)=\{(5,0),(5,1),(5,2).....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

(vi) [a,b] where a\leq b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

[a,b]=\{[5,0],[5,1],[5,2].....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

8 0
3 years ago
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