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goldenfox [79]
3 years ago
12

What does x equal, I'm struggling with this?​

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
5 0

Answer:

85

Step-by-step explanation:

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Solve for x in the diagram below
MrRissso [65]

Answer:

x=20 degrees

Step-by-step explanation:

because this is a supplementary angle, you add everything and set it equal to 180 degrees. so it would be x + 100 + 3x = 180. combine like terms to get 100 + 4x = 180. subtract 100 from both sides to get 4x = 80. divide by 4 on both sides to isolate the x and your final answer is 20. hope this helped! good luck in school xoxo

8 0
3 years ago
i just missed a couple says of school and i need someone to explain how to do this to me: please help
algol [13]
Remember the formula y = mx+b.

'm' in mx is the slope.

'b' is the y-intercept.

So in the equation y = -3x - 3...

The slope is -3. The y-intercept is -3.

I hope you find this answer the most helpful! :)
3 0
3 years ago
Alexis G. has a cell phone plan that charges $0.05 per minute plus a monthly fee of $25.00. She budgets $35.50 per month for tot
jeyben [28]
The Maximum amount of minutes she could use would be 210

Formula:
0.05x + 25 <= 35.50
6 0
3 years ago
Can someone help with step by step
MrMuchimi

Answer:

See explanation

Step-by-step explanation:

In\: \triangle ABC \:\&\: \triangle EDC\\\angle ABC \cong \angle EDC... (Right \: \angle s) \\\angle ACB \cong \angle ECD... (Vertical \: \angle s) \\\therefore \triangle ABC \sim \triangle EDC... (By\: AA\: Postulate) \\

3 0
3 years ago
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

b. I = 2. 6 × 10^-3 A

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

I = V/R

Where v = voltage

R = resistance

A. V = 12V

R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

I = \frac{12}{19700}

I = 6. 1 × 10^-4 A

B. V = 9V

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

I = \frac{12}{4700}

I = 2. 6 × 10^-3 A

C.  V=  12V

1/R = \frac{1}{750} + \frac{1}{1200} + \frac{1}{950 } = 3. 22 × 10^-3

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

I = \frac{12}{310. 56}

I = 0. 04 A

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:

brainly.com/question/14296509

#SPJ1

8 0
2 years ago
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