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2 years ago

Write the equation of the line, in standard form that has an x intercept at (3,0) and y

Mathematics

1 answer:

Nataly [62]2 years ago

6 0

x intercept=a=3
y intercept=b=-2

We need the normal form first

$\\ \rm\bull\rightarrowtail \dfrac{x}{a}+\dfrac{y}{b}=1$

$\\ \rm\bull\rightarrowtail \dfrac{x}{3}-\dfrac{y}{2}=1$

Multiply by 6to bring standard form

$\\ \rm\bull\rightarrowtail 2x-3y=1$

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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

HELP I NEED TO KNOW IN LESS THAN AN HOUR

Vadim26 [7]

Slope=5. The change in y is +5and the change in x values is +1. 5/1=5

8 0

3 years ago

Find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

kherson [118]

Answer:

$=\frac{364\pi}{3}$ cubic units

Step-by-step explanation:

We are to find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

f(x)=2x+1, y=0, x=0, x=4.

The picture is given as shaded region.

This is rotated about x axis

Limits for x are already given as 0 and 4

f(x) is a straight line

The solid formed would be a cone

Volume = $\pi \int\limits^a_b {(2x+1)^2} \, dx \\= \pi \int\limits^4_0 {(4x^2+4x+1)} \, dx \\=\pi [\frac{4x^3}{3} +2x^2+x]^5_0\\\\=\pi[\frac{4*4^3}{3}+2*4^2+4-0]\\=\frac{364\pi}{3}$

7 0

2 years ago

John can ride his bicycle at a rate of 14 miles per hour. Convert this rate to feet per second_

azamat

Answer:

Step-by-step explanation:

Givens

r = 14 miles / hour

Problem

Convert to feet per second.

Solution

Later on, you will learn how to solve these kinds of conversion problems using a method called Unit Analysis. Since you likely have not been taught this method, will just use proportions.

1 mile = 5280 feet

14 miles = x Cross multiply

x = 14 * 5280 Simplify

x = 73920 feet

1 hour = 3600 seconds

Answer

r = 73920 feet / 3600 seconds

r = 20.533 feet / secpmd

7 0

1 year ago

James deposits $400 in a bank account for three years. He will earn 2.5% compound interest per annum. How much will be in his ba

IceJOKER [234]

Answer:

i think its $1000 after 3 year

4 0

3 years ago