2 years ago

22x31 a537 b682 c492

Mathematics

1 answer:

Aloiza [94]2 years ago

5 0

Step-by-step explanation:

copy from there that right answer if need more then follow me Bhai or send me money

You might be interested in

Yuri thinks that 3/4 is a root of the following function. q(x) = 6x3 + 19x2 – 15x – 28 Explain to Yuri why 3/4 cannot be a root.

k0ka [10]

Answer:

Yuri is not correct.

Step-by-step explanation:

Given expression is q(x) = 6x³ + 19x² - 15x - 28

If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0

Similarly, for x = $\frac{3}{4}$ ,

$q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28$

= $6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28$

= $(\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28$

= $(\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}$

= $-\frac{1666}{64}$

= $-\frac{833}{32}$ ≠ 0

Therefore, Yuri is not correct. x = $\frac{3}{4}$ can not be a root of the given expression.

6 0

3 years ago

Read 2 more answers

1. x=1/(root3-root2). find rootx-(1/rootx) 2. if x=[root(a+2b)+root(a-2b)]/[root(a+2b)-root(a-2b]. show that bx^2-ax+b=0

timofeeve [1]

1. x = 1/ ( $\sqrt{3} - \sqrt{2)}$ = $\sqrt{3}+ \sqrt{2}$ ;
( $\sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =$ =

6 0

3 years ago

Read 2 more answers

K + 9 = 17 What does k value? k=8 k=9 k=17 k=26

belka [17]

Answer:

Step-by-step explanation:

k + 9 = 17

subtract 9 from both sides:

k = 8

6 0

3 years ago

Read 2 more answers

What is 3^3/2 equal to?

antoniya [11.8K]

Answer:

D. 2^3/2

Explanation:

3 3/2

= ²√3³

= ²√27

Hope this helped! :D

6 0

3 years ago

Read 2 more answers

Bernard solved the equation 5x+(-4)=6x+4 using algebra tiles.Which explains why Bernard added 5 negative x-tiles to both sides i

Volgvan

5x+(-4)-5x=6x+4-5x
-4=x+4
x=-8

To get rid of 5x on the left side.

6 0

3 years ago

Read 2 more answers