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Vaselesa [24]
2 years ago
14

A function in algebra is

Mathematics
2 answers:
fenix001 [56]2 years ago
8 0

Answer:

A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a specified type. It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.

Step-by-step explanation:

im boared so i answered

Elanso [62]2 years ago
5 0
Is an equation that has only one answer for every x
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Drag a vertex of the triangle to change its shape.
likoan [24]

Answer:

The correct option is;

45°

Step-by-step explanation:

By angle sum theorem, we have that the sum of angles in a triangle = 180°

Therefore, we have;

When the interior angles of the triangle are constructed to be 60° and 75°, we have by the angle sum theorem;

The third angle + 60° + 75° = 180°

Which gives;

The third angle  = 180°- 60° - 75° = 180°- 135° = 45°

The measurement of the third angle by the angle sum theorem will be 45°

The correct option is ∠third angle = 45°.

8 0
3 years ago
One-half of the sum of a number and 6 is less than 25. What is the number
Leno4ka [110]

Answer:

44

Step-by-step explanation:

1/2(n + 6) < 25         Distribute

1/2n + 3 < 25

       - 3     - 3           Subtract 3 from both sides

1/2n < 22                  Multiply both sides by 2 or divide by 1/2

n < 44

If this answer is correct, please make me Brainliest!

4 0
3 years ago
A vertical pole 5ft long casts a shadow of 2ft at the same time a nearby tree casts a shadow of 10ft, how y’all is the tree?
SSSSS [86.1K]

Answer: The tree is 4 feet tall.

Step-by-step explanation: 5/2=x/10

2 x 10=20

20/5=4

5 0
3 years ago
Find the sum 1/6 (12C + 24) + 1/3(12- 3)<br><br>a-b-c- or d
Anarel [89]
<span>1/6 (12C + 24) + 1/3(12 c -  3) = 

12 c/ 6  + 4 +12c / 3 - 1 = 

2c +4 + 4 c -1 

6c +3 </span>
7 0
3 years ago
HEY CAN ANYONE HELP ME OUT IN DIS PLS
dimulka [17.4K]
The answer is C.
You have to subtract 13 from both sides.
7 0
3 years ago
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