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ki77a [65]
3 years ago
13

PLSSS HELP DUE IN 5 MINUTES!!!!

Mathematics
2 answers:
MissTica3 years ago
8 0

Answer:

(3, 0)

Step-by-step explanation:

The vertex point is 3 units to the right of the origin, making the x-coordinate 3, but it doesn’t move up or down, meaning that the y-coordinate is 0.

so it’s (3,0)

Btw, the vertex is the point of the parabola where it’s the minimum or maximu (minimum in this case). You answered for the y-interceptm which is where the parabola touches the y-axis.

Pavel [41]3 years ago
4 0

Answer:

I think it's (0,3) sorry if it not

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Ill mark brainlist plss help
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200 mm and 200cm are too small.

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The 6 elephants at the zoo have a combined weight of 11,894 pounds. About how much does each elephant weigh?
Sidana [21]

Answer: 14,000 i think

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2 years ago
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What is the range of the data set below?<br><br> 56, 42, 48, 32, 17, 26, 92
Digiron [165]

Answer:

edit: 92-17=75

Step-by-step explanation:

when using range you subtract the highest number with the lowest

3 0
3 years ago
) Supriya makes papadums to sell in her food truck. She uses 12 oz of dough to make each batch.
Anna11 [10]

The number of batches she can make is = 5 batches

<h3>Calculation of total number of batches</h3>

The amount of dough used to make each batch = 12oz

The total amount of dough she has= 60oz

Therefore, the number of batches she can make from the total dough available = b

Using the equation:

(dough for one batch) × (number of batches) = (total amount of dough)

which is = 12× b = 60

b= 60/12

b= 5 batches

Learn more about batches here:

brainly.com/question/25770607

4 0
2 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height
Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

y(t)=43t-23t^{2}

B) Instantaneous Velocity at t = 1

y(1) = 43(1)-23(1)^2 = 20

A) Formula:

Average velocity =

\displaystyle\frac{\text{Displacement}}{\text{Time}}

1) 0.01

y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}

2) 0.005 s

y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}

3) 0.002 s

y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}

4) 0.001 s

y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}

3 0
3 years ago
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