Find the quotient of (6x^2+x-12)÷(2x+3)

What is the approximate distance from the origin to the point (3, −4, 5)? Round to the nearest unit. 4 units 7 units 8 units 9 u

Dafna11 [192]

Answer:

Option B is correct.

Step-by-step explanation:

The distance formula used is:

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}$

We need to find distance between origin and other point (3,-4,5)

Origin is: (0,0,0)

x₁ = 0, y₁ = 0, z₁ =0 and x₂= 3, y₂= -4 and z₂ = 5

Putting values in the distance formula we get:

$d=\sqrt{(3-0)^2+(-4+0)^2+(5-0)^2}\\d=\sqrt{(3)^2+(-4)^2+(5)^2}\\d=\sqrt{9+16+25}\\d=\sqrt{50}\\d=7$

The Distance from the origin to the point (3, −4, 5) is 7 units.

Option B is correct.

3 0

3 years ago

Find an equation of the sphere containing all surface points P = (x, y, z) such that the distance from P to A(−3, 6, 3) is "twic

Marina86 [1]

Answer:

Equation of Sphere = $x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

Step-by-step explanation:

Data Given:

P = (x,y,z)

Distance from P to A (-3,6,3) = Twice the distance from P to B(6,2,-3)

Solution:

Find the equation of the sphere:

It is given that:

PA = 2PB

Squaring both sides:

$(PA)^{2} = (2PB)^{2}$

$(PA)^{2} = 4 (PB)^{2}$

$(x - (-3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x $(x-6)^{2}$ + $(y-2)^{2}$ + $(z-(-3))^{2}$

Solving the above equation:

$(x + 3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x { $(x-6)^{2}$ + $(y-2)^{2}$ + $(z + 3))^{2}$ }

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 { $x^{2}$ + 36 - 12x + $y^{2}$ + 4 - 4y + $z^{2}$ + 9 + 6z}

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 $x^{2}$ + 144 - 48x +4 $y^{2}$ + 16 - 16y + 4 $z^{2}$ + 36 + 24z

Putting the right hand side = 0 and solving the equation:

$x^{2}$ - 4 $x^{2}$ + $y^{2}$ - 4 $y^{2}$ + $z^{2}$ - 4 $z^{2}$ + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0

-3 $x^{2}$ - 3 $y^{2}$ -3 $z^{2}$ + 54x + 4y -30z -142 = 0

Taking (-) common

- ( 3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142) = 0

3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142 = 0

dividing the whole equation by 3

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

7 0

3 years ago

Which of the diagrams below represents the contrapositive of the statement "If it is an elm, then it is a tree"?

Murrr4er [49]

Answer:

Right

Step-by-step explanation:

For a contrapositive, we switch the hypothesis and conclusion and then negate it.

Switch

If it is a tree, then it is an elm

Negate

If it is not a tree, then it is not an elm.

Not tree belongs inside not elm

Right

7 0

3 years ago

Check whether 2-3x is a factor of (x)2 -x+1

sergejj [24]

Answer:

no

Step-by-step explanation:

if it was a factor then, 2-3x = 0

x= 2/3

putting x = 2/3 in eqn,

(2/3)^2 - 2/3 +1 =7/9....so it is not a factor

5 0

3 years ago

Read 2 more answers

Verifying Trig Functions

anzhelika [568]

Answer:

The answer to your question is below

Step-by-step explanation:

csc α - cot α = $\frac{sin \alpha }{1 + cos \alpha }$

Work with the left part

$\frac{1}{sin\alpha } - \frac{cos \alpha }{sin \alpha}$

Simplify

$\frac{1 - cos\alpha }{sin\alpha }$

Multiply by the reciprocal

$\frac{1 - cos\alpha }{sin \alpha } x \frac{1 + cos\alpha }{1 + cos\alpha }$

Simplify

$\frac{1 - cos^{2}\alpha}{sin\alpha (1 + cos\alpha)}$

Remember that

sin²α = 1 -cos²α

Substitute the previous equation

$\frac{sin^{2}\alpha }{sin\alpha(1 + cos\alpha ) }$

Simplify

$\frac{sin\alpha}{1 + cos \alpha}$

The identity i proved.

5 0

3 years ago

Find the quotient of (6x^2+x-12)÷(2x+3)​

Find the quotient of (6x^2+x-12)÷(2x+3)