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Tpy6a [65]
3 years ago
11

Order - 11, 22, -10, 17, -26, and 1 from greatest to least.​

Mathematics
1 answer:
Ratling [72]3 years ago
3 0
Greatest 22,17,1,-10,-11,-26 least
You might be interested in
PLEASE HELP!!!!!!!!!!!!!
Anna35 [415]
4y^2+2=(4y\cdot4y)+2\\
4y^2+2=16y^2+2 \leftarrow \text{not identity}\\\\
4y^2+2=(4y+y)+2\\
4y^2+2=5y+2 \leftarrow \text{not identity}\\\\
4y^2+2=4(y+y)+2\\
4y^2+2=8y+2 \leftarrow \text{not identity}\\\\
4y^2+2=4(y\cdot y)+2\\
4y^2+2=4y^2+2 \leftarrow \text{identity}

So it's D.
8 0
3 years ago
What’s the answer to this???
Readme [11.4K]

Answer:

5+6-8

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What is the answer to <br> 6+0.10x=0.15x+8
asambeis [7]

Let's solve this problem step-by-step.

6+0.1x=0.15x+8

Step 1: Simplify both sides of the equation.

0.1x+6=0.15x+8

Step 2: Subtract 0.15x from both sides.

0.1x+6−0.15x=0.15x+8−0.15x

−0.05x+6=8

Step 3: Subtract 6 from both sides.

−0.05x+6−6=8−6

−0.05x=2

Step 4: Divide both sides by -0.05.

-0.05x/-0.05=2/-0.05

So, the answer for this problem is x=-40

4 0
3 years ago
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
If a line goes through the point (8,9) and has a slope of negative 3/4 what is the starting value?
scZoUnD [109]

I assume by starting value, you mean y-intercept. The y-intercept is 15

To find this, we need to use slope intercept form and the given information to find it.

y = mx + b ----> plug in the known values

9 = -3/4(8) + b

9 = -6 + b

15 = b

Therefore, the starting value is 15.

5 0
3 years ago
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