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3 years ago

Write the equation in point-slope form of the line that passes through the given points.

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

4 0

Answer:

y-6=-1/3(x+6)

Step-by-step explanation:

y-y1=m(x-x1)

m=(y2-y1)/(x2-x1)

m=(3-6)/(3-(-6))

m=-3/(3+6)

m=-3/9

simplify

m=-1/3

y-6=-1/3(x-(-6))

y-6=-1/3(x+6)

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Solve for x. x= In e x=

IgorLugansk [536]

Good morning ☕️

Answer:

x = 1

Step-by-step explanation:

since the function ln is the inverse of the function exponential then :

$ln(e )=lne^{1} = 1$

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2 years ago

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3 years ago

The cake store is having a 25% off sale on all of its cakes.

Shalnov [3]

Answer:

$6.75

Step-by-step explanation:

First you find 25% of $9----> 2.25 and then you subtract that from 9----> 9-2.25=6.75

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2 years ago

Read 2 more answers

Pick any number and add 5 to it. Find the sum of the new number and the original number. Add 5 to the sum. Divide the new sum by

Oksana_A [137]

Answer: The final number is 5

Step-by-step explanation:

Picked number: 10

STEP are all according to what you stated

10+5=15 (first sentence)

15+10=25 (second sentence)

25+5=30 (third sentence)

30÷2-10=5 (fourth sentence)

The final number is 5.

3 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago