Ask question

Ask question

All categories

sergiy2304 [10]

3 years ago

7

How do you get the answer

1 answer:

I am Lyosha [343]3 years ago

5 0

Plug in -2 for a, 3 for b, and -5 for c. Then it'd be | (-3)^2 - 2(-2)(-5) +5 (3) |. Then, multiply everything to get | 9 -20 +15 |. Then, add/subtract to get | 4 |. Finally, take the absolute value and get 4.

You might be interested in

What is the equation in point−slope form of the line passing through (2, 2) and (4, 8)?

MrMuchimi

The answer issss D
I hope this helped u :D

3 0

2 years ago

Mr. Potter's class will take a test for 1 1/2 hours. They need to finish by 11:40 to get to lunch on time.What is the latest tim

stich3 [128]

The answer is 10:10

3 0

3 years ago

Please help me mane fr

VMariaS [17]

Hi There!

Answer:

25x² - 4

Step-by-step explanation:

Solve: (5x - 2)²

Step 1: 25x² - 2²

Step 2: 25x² - 4

Hope This Helps :)

7 0

3 years ago

Solve the quadratic by factoring.<br> x^2-5x-31=5

torisob [31]

Answer:

$x = - 4\: or \: x = 9$

Step-by-step explanation:

make one side equal to zero:

${x}^{2} - 5x - 36 = 0$

find two factors that multiply to -36 and add to -5 (-9 & 4)

${x }^{2} - 9x + 4x - 36 = 0$

factorise

$x(x - 9) + 4(x - 9) = 0$

$(x + 4)(x - 9) = 0$

solve for x

$x = - 4 \: or \: x = 9$

5 0

3 years ago

Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.

alina1380 [7]

Answer:

The answer is $x=a^{-1}cb^{-1}$ .

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$ , where $e$ stands for the identity element of G.

So, given the equality $axb=c$ we make a left multiplication by $a^{-1}$ and we obtain:

$a^{-1}axb =a^{-1}c.$

But, $a^{-1}axb = exb = xb$ . Hence, $xb = a^{-1}c$ .

Now, in the equality $xb = a^{-1}c$ we make a right multiplication by $b^{-1}$ , and we obtain

$xbb^{-1} = a^{-1}cb^{-1}$ .

Recall that $bb^{-1}=e$ and $xe=x$ . Therefore,

$x=a^{-1}cb^{-1}$ .

6 0

3 years ago