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Sav [38]
3 years ago
10

Which function has the greater rate of change?

Mathematics
1 answer:
kompoz [17]3 years ago
8 0

Answer: Function 1

Step-by-step explanation:

Function 1 has rate of change of

(15.75 - 14.25)/(1 -(-1)) = 0.75. This was calculated using slope formula.

Function 2 has rate of change of 5/6, which is 0.833.

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A one year member ship to your local gym costs $539. there is an initial fee of $35. write an equation tou can use to fund out h
stich3 [128]
$539 - $35 = $504

$504 : 12 = $42

Per month you pay $42.
4 0
2 years ago
Please help with question 2, I’m too tired to think right now, and it’s for my little sister and it’s due tomorrow! Ignore the n
marishachu [46]

Answer:

2000 L

Step-by-step explanation:

There are 1250 L of water in a tank at present. If the tank is 0.625 full, what is the capacity of the tank?

The simple solution is:

1250 L ÷ 0.625 = 2000 L

The algebraic solution is:

Let <em>c</em> equal the capacity of the tank.

Therefore, <em>c</em> × 0.625 = 1250.

Divide both sides by 0.625:

<em>c</em> × 0.625 ÷ 0.625 = 1250 ÷ 0.625

And simplify:

<em>c</em> = 1250 ÷ 0.625

<em>c</em> = 2000

5 0
3 years ago
Please help answer these questions. My teacher said they were really easy but I just don't understand. Will mark brainliest !!!
kodGreya [7K]

Answer:

1.  A = 59

2.  A = 43

Step-by-step explanation:

If we have a right triangle  we can use sin, cos and tan.

sin = opp/ hypotenuse

cos= adjacent/ hypotenuse

tan = opposite/ adjacent


For the first problem, we know the opposite and adjacent sides to angle A

tan A = opposite/ adjacent

tan A = 8.8 / 5.2

Take the inverse of each side

tan ^-1 tan A = tan ^-1 (8.8/5.2)

A = 59.42077313

To the nearest degree

A = 59 degrees


For the second problem, we know the  adjacent side and the hypotenuse to angle A

cos A = adjacent/hypotenuse

cos A = 15.3/21

Take the inverse of each side

cos ^-1 cos A = cos ^-1 (15.3/21)

A = 43.23323481

To the nearest degree

A = 43 degrees


6 0
3 years ago
What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 4) ​, (2, 7) , ​ (6, 4) ​, and (6, 1) ​?
alekssr [168]
First we need the distances of the sides of the polygon, because perimeter = sum of all sides.
d =  \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }
d1 =  \sqrt{{(1 - 4)}^{2} + {( - 2 -  - 2)}^{2} } \\  =  \sqrt{{(- 3)}^{2} + {(0)}^{2} } =  \sqrt{9}  \: = 3
d2 = \sqrt{{(4 - 7)}^{2} + {( - 2 - 2)}^{2} } \\  = \sqrt{{(- 3)}^{2} + {( -4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25}  \: = 5
d3 \: = \sqrt{{(7 - 4)}^{2} + {(2 - 6)}^{2} } \\  =  \sqrt{{(3)}^{2} +  {( - 4)}^{2} } =  \sqrt{9 + 16}  \\  =  \sqrt{25} \:  = 5
d4 = \sqrt{{(4 - 1)}^{2} + {(6 - 6)}^{2} } \\  =  \sqrt{ {(3)}^{2} +  {(0)}^{2}  }  =  \sqrt{9} \:  = 3
d5 = \sqrt{{(1 - 1)}^{2} + {(-2 - 6)}^{2} }  \\ =  \sqrt{ {(0)}^{2} +  {( - 4)}^{2}} =  \sqrt{16}  \:  = 4
Now, we add all sides for the perimeter:
p = d1 + d2 + d3 + d4 + d5
p = 3+5+5+3+4 = 20 units





5 0
3 years ago
Read 2 more answers
Suppose you add or subtract two quadratic trinomials that use the same variable. What are the possible classifications for the s
Leno4ka [110]

Answer:

quadratic monomial, quadratic trinomial, constant monomial, linear monomial, quadratic binomial, linear binomial.

Step-by-step explanation:

two quadratic trinomials each have a degree-2, degree-1, and degree-0 term. It is possible that the coefficients of all or some of the terms cancel each other while adding or subtracting the polynomials.

3 0
3 years ago
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