2+5+2) = 6(+2)

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

4 years ago

Tara has two standard, six-sided dice. If she rolls them 36 times, how many times should she expect a sum less than 5 to show up

andreyandreev [35.5K]

There are six sides on each die. For each possible score on Die 1, there are six possible scores on Die 2. So the number of possible combinations is 6*6 = 36.
<span>It follows that if the dice are thrown 36 times, you would expect each combination to come up once. </span>

<span>We therefore simply need to know how many combinations add up to less than 5. (I've interpreted this as not including a total of 5 itself). </span>
<span>These combinations are: 1 and 1, 2 and 1, 1 and 2, 2 and 2, 3 and 1, and 1 and 3 ---> six combinations out of 36. </span>

<span>So you'd expect a sum less than 5 six times. </span>

8 0

3 years ago

A airplane flies 370 miles from point A to point B with a bearing of 24

poizon [28]

Hiyoigbv jvgkk uoiu jvduj

4 0

3 years ago

The table shows the last holiday destination of 60 people

Licemer1 [7]

Answer:

Use your protractor to draw the lines

Step-by-step explanation:

You worked out the angles, so using the centre point as the middle and the line underneath France as your starter line, do 108° sector for Spain, then a 90° for Greece and a 12° for Others

3 0

2 years ago

D is the midpoint of CE. If CD= 2x+10 and DE =3x-18, find x and CE<br><br> x=<br> CE=

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

|<------------- CE -------------------->|

C-------------------D-------------------E

2x + 10 3x - 18

2x + 10 = 3x - 18

2x - 3x = -18 - 10

-x = -28

x = 28

CE = 2x + 10 + 3x - 18

CE = 2(28) + 10 + 3(28) - 18

CE = 56 + 10 + 84 - 18

CE = 132

3 0

4 years ago