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Blizzard [7]
2 years ago
9

What is another way to name angle 3?

Mathematics
1 answer:
fiasKO [112]2 years ago
7 0

Answer:

intimately there are three types of angles a cute angle and angle between zero and 90° right angle and angle 90 degree angle of two angle and angle between 90 and 180 degree

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What is the area of the composite figure ​
tigry1 [53]

Answer:

132cm

Step-by-step explanation:

4 0
3 years ago
Given<br> the<br> function<br> f(x)= 3(x-2)² + 5, what<br> is<br> f(-1)?
motikmotik

Answer:

32

Step-by-step explanation:

f(x)= 3(x-2)² + 5

Let x = -1

f(-1)= 3(-1-2)² + 5

Parentheses first

f(-1) = 3(-3)^2 +5

Exponents

f(-1)= 3*9 + 5

Multiply

f(-1) = 27+5

Add

f(-1) = 32

5 0
2 years ago
Read 2 more answers
Need help please help me
ANTONII [103]
Megan:
x to the one third power =  x ^{1/3}
<span>x to the one twelfth power = </span>x ^{1/12}

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>\frac{x ^{1/3}}{x ^{1/12}}
<span>
Since </span>\frac{ x^{a} }{ x^{b} } = x ^{a-b}
then \frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12}

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

\frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12} = x^{1/4}


Julie:
x times x to the second times x to the fifth = x * x² * x⁵

<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>\sqrt[32]{x* x^{2} * x^{5} }
<span>
Since </span>x^{a}* x^{b}= x^{a+b}
Then \sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }

Since \sqrt[n]{x^{m}} = x^{m/n} }
Then \sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}

Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
3 0
3 years ago
In a competition, a school awarded medals in different categiories.40 medals in sport 25 medals in danceand 212 medals in music,
aleksley [76]

Answer:

210

Step-by-step explanation:

Given:

Medals in sports = 40

Medals in dance = 25

Medals in music = 212

Total students that received medals = 55

Total students that received medals in all three categories = 6

Required:

How many students get medals in exactly two of these categories?

Take the following:

A = set of persons who got medals in sports.

B = set of persons who got medals in dance

C = set of persons who got medals in music.

Therefore,

n(A) = 40

n(B) = 25

n(C) = 212

n(A∪B∪C)= 55

n(A∩B∩C)= 6

To find how many students get medals in exactly two of these categories, we have:

n(A∩B) + n(B∩C) + n(A∩C) −3*n(A∩B∩C)

=n(A∩B) + n(B∩C) + n(A∩C) −3*6 ……............... (1)

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)

Thus, n(A∩B)+n(B∩C)+n(A∩C)=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)

Using equation 1:

=n(A)+n(B)+n(C)+n(A∩B∩C)−n(A∪B∪C)−18

Substitute values in the equation:

= 40 + 25 + 212 + 6 − 55 − 18

= 283 - 73

= 210

Number of students that get medals in exactly two of these categories are 210

6 0
4 years ago
How do I do long devision
Marina CMI [18]

Answer:

basicly you divided muiltiply and subtract bring back down then bring it on back

6 0
3 years ago
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