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Natalija [7]
3 years ago
12

Solve the system of equations using substitution. Show your work. y= 2x - 10 y= 4x - 8

Mathematics
1 answer:
Semenov [28]3 years ago
4 0
Y=2x-10
Y=4x-8
U substitute it and it turns into 2x-10 =4x-8
U get x=-1
Y=4*(-1)-8
Y=-12
(x,y)= (-1,-12)
If u have a phone u can use photo math or if u don’t there’s a thing called mathpapa
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True or False: The slope of the line that passes through (9, 9) and (6, 4) is 3/5.
emmainna [20.7K]

Answer:

False.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

Point (9, 9)

Point (6, 4)

<u>Step 2: Find slope </u><em><u>m</u></em>

  1. Substitute:                     m=\frac{4-9}{6-9}
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HELPP 35 POINTS!!!!! Point Q is the center of Circle Q in the diagram below. The measure of angle XAY is 72° (m∠XAY = 72°). Sinc
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Answer:

Se explanation

Step-by-step explanation:

The diagram shows the circle with center Q. In this circle, angle XAY is inscribed angle subtended on the arc XY. Angle XQY is the central angle subtended on the same arc XY.

The inscribed angle theorem states that an angle inscribed in a circle is half of the central angle that subtends the same arc on the circle. Therefore,

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The measure of the intercepted arc XY is the measure of the central angle XQY and is equal to 144°.

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Read 2 more answers
Determine the current through each of the LEDs in the circuits below. Which LED will be
Anika [276]

a. I = 6. 1 × 10^-4 A

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c. I = 0. 04 A

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

<h3>How to determine the current</h3>

The formula for finding current

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Where v = voltage

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A. V = 12V

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I = 6. 1 × 10^-4 A

B. V = 9V

R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series

I = \frac{12}{4700}

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R = \frac{1}{3. 22 * 10 ^-3} = 310. 56 Ω

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It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

Learn more about Ohms law here:

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#SPJ1

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