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Sophie [7]
3 years ago
8

What are two different ways 23x=10 could be solved for x?

Mathematics
1 answer:
lora16 [44]3 years ago
4 0

Answer:

3 & 4

Step-by-step explanation:

Given

\frac{2}{3}x = 10

Required

Determine the number of ways to solve for x

From the list of given options, (1), (2) & (5) are incorrect, as they won't give the solution to the value of x

Option (3) is correct: See below

\frac{2}{3}x/\frac{2}{3} = 10/\frac{2}{3}

\frac{2}{3}x*\frac{3}{2} = 10 * \frac{3}{2}

x = 10 * \frac{3}{2}

x = \frac{30}{2}

x = 15

Option (4) is also correct: See below

\frac{2}{3}x*\frac{3}{2} = 10 * \frac{3}{2}

x = 10 * \frac{3}{2}

x = \frac{30}{2}

x = 15

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Answer:

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3 years ago
gal created a painting with an area of 56 square inches and a length if 7 inches. they create a second painting with an area of
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5 inches

Step-by-step explanation:

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2 years ago
why do we need imaginary numbers?explain how can we expand (a+ib)^5. finally provide the expanded solution of (a+ib)^5.(write a
zheka24 [161]

Answer:

a. We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution.

b. (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

Step-by-step explanation:

a. Why do we need imaginary numbers?

We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution. For example, the equation of the form x² + 2x + 1 = 0 has the solution (x - 1)(x + 1) = 0 , x = 1 twice. The equation x² + 1 = 0 has the solution x² = -1 ⇒ x = √-1. Since we cannot find the square-root of a negative number, the identity i = √-1 was developed to be the solution to the problem of solving quadratic equations which have the square-root of a negative number.

b. Expand (a + ib)⁵

(a + ib)⁵ =  (a + ib)(a + ib)⁴ = (a + ib)(a + ib)²(a + ib)²

(a + ib)² = (a + ib)(a + ib) = a² + 2iab + (ib)² = a² + 2iab - b²

(a + ib)²(a + ib)² = (a² + 2iab - b²)(a² + 2iab - b²)

= a⁴ + 2ia³b - a²b² + 2ia³b + (2iab)² - 2iab³ - a²b² - 2iab³ + b⁴

= a⁴ + 2ia³b - a²b² + 2ia³b - 4a²b² - 2iab³ - a²b² - 2iab³ + b⁴

collecting like terms, we have

= a⁴ + 2ia³b + 2ia³b - a²b² - 4a²b² - a²b² - 2iab³  - 2iab³ + b⁴

= a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴

(a + ib)(a + ib)⁴ = (a + ib)(a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴)

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b + 4i²a³b² - 6ia²b³ - 4i²ab⁴ + ib⁵

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b - 4a³b² - 6ia²b³ + 4ab⁴ + ib⁵

collecting like terms, we have

= a⁵ + 4ia⁴b + ia⁴b - 6a³b² - 4a³b² - 4ia²b³ - 6ia²b³ + ab⁴ + 4ab⁴ + ib⁵

= a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

So, (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

5 0
3 years ago
HELPPPPP MEEEEE PLEASEEEE
Korolek [52]

Answer:

I don't know sooooooooop

mark me as brainliest ok understood

8 0
3 years ago
Read 2 more answers
What is the value of x in this equation?
andre [41]
The value of x is
-4x + 8 = 42
-4x = 42 -8
-4x = 34
x = -8,5
8 0
3 years ago
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