All categories

3 years ago

What is the conjugate 2x^2 + square root of 3

Mathematics

1 answer:

finlep [7]3 years ago

5 0

Answer:2x^2-√(3)

Step-by-step explanation:

Sign change in the conjugate

Therefore we have 2x^2-√(3)

You might be interested in

What is the equation of the horizontal line through (1,9)

stealth61 [152]

Answer:

slope=9

Step-by-step explanation:

the slope is 9, because you do the y/x or rise over run, so you do for every 9 up 1 right to get a slope of 9/1 or 9

3 0

3 years ago

Read 2 more answers

Given: f(x) = 3x-7, g(x)=2x2 -3x+1, h(x)=4x+1, k(x) =-x2+3 Find: k(x) +g(x)

Afina-wow [57]

Answer:

k(x) + g(x) = x² - 3x + 4

General Formulas and Concepts:

Alg I

Combining like terms

Step-by-step explanation:

Step 1: Define

f(x) = 3x - 7

g(x) = 2x² - 3x + 1

h(x) = 4x + 1

k(x) = -x² + 3

Step 2: Find k(x) + g(x)

Substitute: k(x) + g(x) = -x² + 3 + 2x² - 3x + 1
Combine like terms (x²): k(x) + g(x) = x² + 3 - 3x + 1
Combine like terms (Z): k(x) + g(x) = x² - 3x + 4

7 0

3 years ago

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago

Solve the equation and justify the steps used to solve. 4+q=17

frez [133]

The answer is 13 … just minus 17-4 and it will equal 13

8 0

3 years ago

cupoosta [38]

Answer:

-2 15 ...................

8 0

3 years ago

Read 2 more answers