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3 years ago

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three points have coordinates A(-1,2),B(3,10)and C(p,8).Find the value (s) of p if AC is perpendicular to BC.

1 answer:

Reil [10]3 years ago

4 0

Step-by-step explanation:

All steps are in pic above.

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What is the range of the data set?

777dan777 [17]

Answer:

42.9

Step-by-step explanation:

The range is the difference between the highest and lowest numbers

71.8-28.9=42.9

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3 years ago

write a division equation to represent the repeated subtraction 32 take away 8 is 24 24 take away 8 is16 16 take away 8 is 8/8 t

grin007 [14]

What this setup essentially represents is “How many 8s can we take away from 32 before hitting 0?” Which in turn can be reframed as “How many 8s fit into 32?” This can be captured in the expression 32 / 8. As we can see from the problem, we can take away 4 8’s before hitting 0, so that gives us the equation 32 / 8 = 4

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3 years ago

3.16×10 tool the third power

Whitepunk [10]

The answer is 160 to this problem

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3 years ago

If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction

valentina_108 [34]

Answer:

a) $\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$ .

b) $Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}$

Step-by-step explanation:

Given a function $f(x,y,z)$ , this function has the following gradient:

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

(a) find the gradient of f

We have that $f(x,y,z) = x\sin{yz}$ . So

$f_{x}(x,y,z) = \sin{yz}$

$f_{y}(x,y,z) = xz\cos{yz}$

$f_{z}(x,y,z) = xy \cos{yz}$ .

$\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}$ .

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

$\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}$

$\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}$ .

$\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)$

The vector is $v = i + 3j - k = (1,3,-1)$

To use v as an unitary vector, we divide each component of v by the norm of v.

$|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}$

So

$v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})$

Now, we can calculate the scalar product that is the directional derivative.

$Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}$

6 0

3 years ago

How do you combine like terms of -8(r-2) + 6(r-2) ??? Will mark brainliest if the answer is right.

solmaris [256]

Answer:

$-2r+4$

Step-by-step explanation:

$-8(r-2)+6(r-2)$

Distribute:

$(-8)(r)+(-8)(-2)+(6)(r)+(6)(-2)\\-8r+16+6r+-12$

Combine like terms:

$(-8r+6r)+(16+-12)\\-2r+4$

-2r + 4 is your answer.

6 0

3 years ago