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rusak2 [61]
3 years ago
6

PLZZZ LOOK AT THE PHOTO PROVIDE HELPPPP I NEEED HELPPPPPPPPPOPP

Mathematics
2 answers:
svetoff [14.1K]3 years ago
6 0

Answer:

ITS A

trust me it's quite simple. you just need to find the overall pattern of the sequence

hoa [83]3 years ago
5 0

Answer:

OH MY GOSH I DON'T KNOW BUT GL! WISH YOU THE BEST AAAAA-

Step-by-step explanation:

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Help please! I am struggling help would be greatly needed
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the first one with the two lines coming off of 2008 because each y can only have one ex

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3 years ago
Which is the mode of this data set?<br> 55, 78, 43, 39, 78, 61, 75, 50, 43, 78
Vikentia [17]

Answer:

78

Step-by-step explanation:

Mode = number that appears most in a set of numbers

The number that appears the most in this set is 78. So, the mode is 78. Hope it helps!

6 0
3 years ago
Read 2 more answers
HELP WILL GIVE BRANLIEST!
Vesna [10]

Answer:

I'm pretty positive it's All real answers?

7 0
3 years ago
If f(x) = 5x – 2 and g(x) = x+1, find (f = g)(x).
Karo-lina-s [1.5K]

Answer: Choice A) 5x^3 - x - 3

=================================================

Work Shown:

We subtract the two functions like so

(f - g)(x) = f(x) - g(x)

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

(f - g)(x) = 5x^3 - 2 - x - 1

(f - g)(x) = 5x^3 - x - 3 ..... choice A

Note: be sure to remember to distribute the negative to every term inside (x+1), and not just to the x only.

5 0
3 years ago
NEED ANSWER ASAP- MARK AS BRANLIEST
azamat

Answer:

  x = 1 +√5

Step-by-step explanation:

There are different formulas for the area of a triangle available, depending on the given information.

<h3>Formulas</h3>

When two sides and the angle between them are given, the relevant area formula is ...

  Area = 1/2(ab)sin(C)

When the base and height of a triangle are given, the relevant area formula is ...

  Area = 1/2bh

<h3>Equal Areas</h3>

The problem statement tells us the two triangles shown have equal areas. That means the two formulas will give the same result.

  Area from angle = Area from base/height

  1/2(x·x)sin(30°) = 1/2(x-2)(x+1)

  x² = 2(x² -x -2) . . . . . . . . . . . use sin(30°) = 1/2, multiply by 4

  x² -2x -4 = 0 . . . . . . . . subtract x², eliminate parentheses

  (x -1)² = 5 . . . . . . . . . add 4+1 to complete the square

<h3>Value of x</h3>

  x = 1 ± √5 . . . . . . take the square root, add 1

The value of x must be greater than 2 in order for the triangle side lengths to be positive. (x-2 > 0) This means x = 1-√5 is an extraneous solution.

The value of x is 1 +√5.

5 0
3 years ago
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