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Alex_Xolod [135]
3 years ago
11

By completing the square1. x2 + 5x + 6= 0cing​

Mathematics
1 answer:
Hoochie [10]3 years ago
7 0

Answer:

x=-3 and x=-2

Step-by-step explanation:

I wrote the math down

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(3) A boy has 10 pens in a pack. He gives 2 pens to his friend. What fraction of the pack is left? *
Olegator [25]

Answer: \frac{4}{5}

Step-by-step explanation:

10 pens - 2 pens = 8 pens

Write 8 of 10 as a fraction.

You get \frac{8}{10}, right?

Now, simplify 8/10.

You get 4/5.

8 0
3 years ago
The lengths of the sides of a triangle are in the extended ratio 7 ​: 8 ​:9 . The perimeter of the triangle is 96 cm. What are t
PIT_PIT [208]

Answer:

28, 32, 36

Step-by-step explanation:

If you multiply each 7, 8, and 9 by 4 you get 28, 32, and 36. The sum of those numbers is 96.

5 0
3 years ago
G + 4g; g= -3<br>help idek ​
forsale [732]

Answer:

-15

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7 0
3 years ago
Read 2 more answers
Answer quick please.
olchik [2.2K]

5/2 = 2.5

4/16 = 0.25

2.5/0.25 = 10

Your answer would be 10

hope it helps!

6 0
4 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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