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Alexus [3.1K]
3 years ago
9

NEED HELP!! #2-9 or what ever questions u wanna do please in need of help!

Mathematics
1 answer:
kykrilka [37]3 years ago
8 0
6. X=8cm
This is because the length on the smaller triangle(4cm) needs to be doubled to fit the length of larger triangle,which will be 8cm (4 x 2 = 8)
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(13x2+9x-4)-(12x2-2x+6)<br> add or subtract the polynomials
bearhunter [10]

Answer:

x^{2} +11x-10

Step-by-step explanation:

(13x^{2} +9x-4)-(12x^{2} -2x+6)

Remove the brackets and multiply the minus sign through all the numbers in the second polynomial

13x^{2} +9x-4-12x^{2} +2x-6

Rearrange in the highest order

13x^{2} -12x^{2} +9x+2x-4-6

Simplify

x^{2} +11x-10

6 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
(X-2/3)^2+y^2 =36 <br> What is its center? <br> What is its radius
Hatshy [7]

Answer:

0(2/3; 0)

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
How many dimensions does a polygon have?
Verdich [7]
A polygon is a 2-dimension figure....
3 0
3 years ago
Read 2 more answers
Simplify: square root 27x12 over 300x8
alex41 [277]

Answer:

10

-------------

3x^2

Step-by-step explanation:

First lets simplify what is inside the square root

300x8

---------------

27x12

3 * 100 x^8

------------------

3 *9 x^12

The 3's cancel  and we know a^b/ a^c = a^(b-c)

100/9 x^ (8-12)

100/9 x^-4

sqrt( 100/9 x^-4)

Put the negative exponent in the denominator and make it positive

sqrt( 100 / (9x^4) )

We know that sqrt( a / b) = sqrt(a) / sqrt( b)

sqrt(100)

----------------

sqrt( 9x^4)

10

-------------

3x^2

Read more on Brainly.com - brainly.com/question/12900964#readmore

5 0
3 years ago
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