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3 years ago

I need help answering an equation in standard form y-4 =0

Mathematics

2 answers:

MissTica3 years ago

6 0

I think that equation might have no solution

ludmilkaskok [199]3 years ago

5 0

Standard Form = NA x 10 ^ -1

so... their is no answer to that equation.

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Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps

Firdavs [7]

ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the

x²y = -4 ... (I)

Condition 2: Real parts are the same

x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

x²y = -4 ... (I)
x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

y = -3 - x² ... (II)

Substituting into equation (I)

x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i

x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i

They result in conjugates

4 0

3 years ago

Read 2 more answers

Tell me which 4 letters are right! A? B? C? D? F? E? I WILL GIVE BRAINLIEST

Helen [10]

Answer:

Everything but A and C

Step-by-step explanation:

5 0

2 years ago

Anyone knows the answer?

SIZIF [17.4K]

Asa. angle side angle.

7 0

3 years ago

The box plots show the weights, in pounds, of the dogs in two different animal shelters.

zimovet [89]

*see attachment below for the box plots that is being referred to

Answer:

*The median weight for shelter A is greater than that for shelter B

*The data for shelter B are a symmetric data set

Step-by-step explanation:

Let's take each stated option and examine the box plot to see which is TRUE or NOT.

OPTION 1: "The median weight for shelter A is greater than that for shelter B."

The median for shelter A = 21,

Median for shelter B = 18

Median weight for shelter A (21) is greater than that for shelter B.

This statement in this option is TRUE.

OPTION 2: "The median weight for shelter B is greater than that for shelter A."

Median for B (18) < median of A (21)

This statement in option is NOT TRUE.

OPTION 3: "The data for shelter A are a symmetric data set."

This is NOT TRUE because the length of both whiskers are unequal and also, the box is not divided equally. One side is longer than the other.

OPTION 4: "The data for shelter B are a symmetric data set."

This is TRUE because the length of both whiskers are relatively equal, and also, the rectangular box is divided equally.

OPTION 5: "The interquartile range for shelter A is 8"

This is NOT TRUE.

IQR = Q3-Q1

IQR for shelter A is approximately = 28-17 = 11

5 0

3 years ago

PLEASEE HELP!! Find the area of the kite.

aev [14]

Answer:

Step-by-step explanation:

Area of kite is Area of triangles. Area is 6+3=9

6 0

3 years ago