Given that A = (4,3,-4), B = (2,8,-5) and C = (-5,5,-8), find the vector from A to the midpoint of BC.

1 answer:

6 0

Mid-point of BC, M=(2+(-5),8+5,-5+(-8))/2=(-1.5, 6.5, -6.5)
Vector from A to M = M-A=<(-1.5,6.5,-6.5)-(4,3,-4)>=<-11/2,7/2,-5/2>

You might be interested in

4.Find the measure of arc DF in the figure.

Vladimir79 [104]

Answer:

129 degrees

Step-by-step explanation:

Ok, DAE, as given, is 90 degrees. So, BAD would equal 90 as well, so BAC would be 90-51, 39. And because FAE is vertical angle to BAC, FAE is also 39. So 39 + 90 is 129.

7 0

2 years ago

What is 83023007 in standard form

S_A_V [24]

Answer:

8.3023007 x 10 $^{7}$

Step-by-step explanation:

The reason for this is because we move the decimal places left 7 digits, that is why the exponent is 7.

8 0

3 years ago

Let $x$ be the positive number such that $2x^2 = 4x + 9.$

loris [4]

Solve for x. I'll do it by completing the square:

2x² = 4x + 9

2x² - 4x = 9

x² - 2x = 9/2

x² - 2x + 1 = 11/2

(x - 1)² = 11/2

x - 1 = ± √(11/2)

x = 1 ± √(11/2)

x = 1 ± √22/2

x = (2 ± √22)/2

Take the solution with the positive square root. Then it follows that a = c = 2 and b = 22. So a + b + c = 26.

4 0

2 years ago

Find all rational zeros of the polynomial

rjkz [21]

$x^3+x^2-5x+3=x^3+3x^2-2x^2-6x+x+3\\\\=(x^3-2x^2+x)+(3x^2-6x+3)=x(x^2-2x+1)+3(x^2-2x+1)\\\\=(x^2-2x+1)(x+3)=\underbrace{(x^2-2(x)(1)+1^2)}_{(a-b)^2=a^2-2ab+b^2}(x+3)=(x-1)^2(x+3)\\\\P(x)=0\iff(x-1)^2(x+3)=0\iff(x-1)^2=0\ \vee\ (x+3)=0\\\\x-1=0\ \vee\ x-1=0 \vee\ x+3=0\\\\x=1\ \vee\ x=1\ \vee\ x=-3\\\\Answer:\ \boxed{x_1=1,\ x_2=1,\ x_3=3}$

4 0

3 years ago

Which of the following sets of ordered pairs represents a function?

Zinaida [17]

Shouldn't the answer be: (-1,4),(1,4),(2,4),(-2,4)

8 0

3 years ago

Read 2 more answers