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HACTEHA [7]
2 years ago
7

What is the name of an angle that is formed by one side of a triangle and the extension of an adjacent side?

Mathematics
2 answers:
nevsk [136]2 years ago
8 0
The answer would be ''a''. Exterior angle.
Harlamova29_29 [7]2 years ago
5 0

we know that

Exterior angle:

An exterior angle of a triangle is an angle formed by one side of the triangle and the extension of an adjacent side of the triangle

we can draw diagram

so,

option-A.........Answer

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When Vince does 14 push-ups and 12 sit-ups, it takes a total of 52 seconds. In comparison, he needs 48 seconds to do 12 push-ups
marshall27 [118]

Answer:

it takes 2 seconds to do one situp

it takes 2 seconds to do one pushup

to do 14 pushups, it takes 28 seconds.

to do 12 situps it takes 24 seconds

to do 12 pushups, it takes 24 seconds

BRAINLIEST PLEASE!

hope this helps :)

Step-by-step explanation:

14x + 12y = 52

12x + 12y = 48

^^^ this is an inequality that we can make.  

x stands for pushups

y stands for situps

if we subtract the second equation from the first, we get

2x = 4

x = 2

now we can place x in one of the equations to solve for y

12(2)+12y=48

24 + 12y = 48

12y = 24

y = 2

5 0
2 years ago
Exponents how do you do them
konstantin123 [22]
 3 to the power of 5 
 
 you multiply 3 times 5 instead of 3 times 3 times 3 times 3  times 3  
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7 0
2 years ago
Read 2 more answers
Which of the following functions have a vertical asymptote for values of Ø such that cos Ø = 1?Select all that apply.
igomit [66]

Answer:

y=\cot \theta and y=\csc \theta

Step-by-step explanation:

Note that if \cos \theta=1, then \sin \theta=0.

Functions y=\cos \theta,\ \ y=\sin \theta do not have vertical asymptotes at all.

Vertical asymptotes have functions y=\tan \theta,\ \ y=\cot \theta,\ \ y=\sec \theta,\ \ y=\csc \theta.

Functions y=\tan \theta and y=\sec \theta have the same vertical asymptotes (when \cos \theta =0).

Functions y=\cot \theta and y=\csc \theta have the same vertical asymptotes (when \sin \theta =0). See attached diagram

3 0
3 years ago
Simplify:<br><br><br> please and ty
bagirrra123 [75]
32 i hope this helps :))
5 0
3 years ago
Can u guys pls help me with this homework
gladu [14]

The value of √(7 * 23 - 1)/8 is 4.47, the values of a, b and c are -14/11, -10/11 and 3, respectively and the area of the shape is 5√5 + 5 square meters

<h3>How to evaluate the radical expression?</h3>

The question goes thus:

If √5 = 2.236, evaluate √(7 * 23 - 1)/8

We have:

√(7 * 23 - 1)/8

Evaluate the product of 7 and 23

√(7 * 23 - 1)/8 = √(161 - 1)/8

Evaluate the difference of 161 and 1

√(7 * 23 - 1)/8 = √160/8

Evaluate the quotient of 160 and 8

√(7 * 23 - 1)/8 = √20

Express 20 as the product 4 and 5

√(7 * 23 - 1)/8 = √(4 * 5)

Expand the product

√(7 * 23 - 1)/8 = √4 * √5

Express √4 as 2

√(7 * 23 - 1)/8 = 2 * √5

Substitute √5 = 2.236

√(7 * 23 - 1)/8 = 2 * 2.236

Evaluate the product

√(7 * 23 - 1)/8 = 4.472

Approximate

√(7 * 23 - 1)/8 = 4.47

Hence, the value of √(7 * 23 - 1)/8 is 4.47

<h3>How to simplify the radical expression?</h3>

The expression is given as:

(3√2 + 5√6)/(3√2 - 5√6)

Rationalize the above expression

(3√2 + 5√6)/(3√2 - 5√6) * (3√2 + 5√6)/(3√2 + 5√6)

Evaluate the product

(3√2 + 5√6)²/((3√2)² - (5√6)²)

Simplify the denominator

(3√2 + 5√6)²/(18 - 150)

This gives

[(3√2)² + (5√6)² + 2 *(3√2) * (5√6)]/(-132)

Simplify the numerator

[168 + 120√3]/(-132)

Simplify the fraction

-14/11 - 10√3/11

Hence, the values of a, b and c are -14/11, -10/11 and 3, respectively

<h3>How to determine the area?</h3>

The area is calculated as:

A = 1/2 * (Sum of parallel bases) * Height

So, we have:

A = 1/2 * (4 + 3√5 + 6 - √5) * √5

Evaluate the like terms

A = 1/2 * (10 + 2√5) * √5

Evaluate the product

A = (5 + √5) * √5

Evaluate the product

A = 5√5 + 5

Hence, the area of the shape is 5√5 + 5 square meters

Read more about rational expressions at:

brainly.com/question/8008240

#SPJ1

6 0
1 year ago
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