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sveta [45]
3 years ago
12

Georgianna wants to use the linear model associated with the data in the table to make a prediction. Which range of time values

describes the entire interval over which she would be interpolating? 0 to 5 minutes 0 to 18 minutes 0 to 30 minutes 18 to 30 minutes
Mathematics
2 answers:
aniked [119]3 years ago
8 0
It should be 0 to 30 minutes.
Lesechka [4]3 years ago
5 0
The answer is 0 to 30 minutes :)
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Someone bought three t-shirts for $15 and nine $0. 90 each including tax how much did he spend
Goshia [24]

You multiply three( # of shirts) and 15.90 ( price)

The answer is $47.70 is how much he spend

6 0
3 years ago
A: x/4 +1 = -3 B: x+4 = -12 1) How can we get Equation B from Equation A?
Sveta_85 [38]

Answer:In the equation y = mx + b for a straight line, the number m is called the slope of the line. Let x = 0, then y = m • 0 + b, so y = b. The number b is the coordinate on the y-axis where the graph crosses the y-axi

Step-by-step explanation:

5 0
3 years ago
I need help with number 16 please.
Kruka [31]
The graph is misleading because the sizes compared to the percentages are off. a more appropriate way to show the data would be with a bar graph.
Hope this helps!
Vote me brainliest!
8 0
3 years ago
Help <br> 50 points and brainleist
UkoKoshka [18]

Answer: -0.5c

Step-by-step explanation: multiply -2.5 by 1/5 (0.2) and you get -0/5

3 0
2 years ago
Read 2 more answers
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
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