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3 years ago

Write an equation in slope-intercept form for the line that passes through (0,1) and (1,3)

Mathematics

1 answer:

Lana71 [14]3 years ago

7 0

Answer:

y= 2x+1

Step-by-step explanation:

Points:

(0,1) and (1,3)

Form of the line:

y=mx+b, m- the slope, b- y-intercept

Finding the slope:

m= (y2-y1)/(x2-x1)
m=(3-1)/(1-0)= 2/1= 2

Line is now:

y= 2x+b

Using one of the given points to find out the value of b:

1=2*0+b
b=1

So the equation for the line is:

y= 2x+1

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3 years ago

How many bananas are needed to make the third scale balance?

frozen [14]

The number of bananas required to make the third scale balance is 3.

<h3>What is a scale?</h3>

A scale is a tool used for weighting objects. If the objects on each side weigh the same the scale is balanced. Based on this, let's determine the possible weight of each fruit to balance the last scale.

Scale 1 (10 bananas/ 2 pineapples)

It is known 10 bananas weigh as much as 2 apples, but to solve this problem let's assume 10 bananas weigh a total of 1000 grams.

10 banans = 1000 grams = each banana weighs 100 grams
2 pineapples = 1000 grams = each pineapple weighs 500 grams

Scale 2 ( 1 pineapple / 2 bananas, 1 apple)

Using the same imaginary weights let's find the weight of the apple.

1 pineapple = 500 grams
2 bananas = 200 grams
1 apple = 300 grams (500 - 200 = 300)

Scale 3 (1 apple /x )

We know 1 apple is equal to 300 grams and 1 banana is equal to 100 grams. Based on this, the number of bananas to balance the scale is 3.

Learn more about balance in: brainly.com/question/7181548

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3 years ago

find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in R3 that ar

Leona [35]

$\mathbf a=(1,1,1)^\top=1\,\mathbf i+1\,\mathbf j+1\,\mathbf k$
$\mathbf b=(-2,3,0)^\top=-2\,\mathbf i+3\,\mathbf j$
$\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\-2&3&0\end{vmatrix}=-3\,\mathbf i-2\,\mathbf j+5\,\mathbf k=(-3,-2,5)^\top$

Basically, you're looking for a matrix $\mathbf A$ such that

$\mathbf A(\mathbf a\times\mathbf b)=\mathbf 0$

i.e. a matrix $\mathbf A$ whose nullspace with basis vector $\mathbf a\times\mathbf b$ .

By the rank-nullity theorem, the rank of $\mathbf A$ and the dimension of its nullspace must add up to the number of columns, so

$\mathrm{rank}\mathbf A+\underbrace{\mathrm{null}\mathbf A}_1=3\implies\mathrm{rank}\mathbf A=2$

One easy choice for a row would be $\begin{bmatrix}1&1&1\end{bmatrix}$ , since

$(1,1,1)(-3,-2,5)^\top=0$

Now you only need to find another combination such that the second row of $\mathbf A$ is independent of the first. An easy choice for this is to let the first element be 0, and the next be 1. Then the last element must be $\dfrac25$ , as

$\left(0,1,\dfrac25\right)(-3,-2,5)^\top=0$

So,

$\underbrace{\begin{bmatrix}1&1&1\\0&1&\frac25\end{bmatrix}}_{\mathbf A}\begin{bmatrix}-3\\-2\\5\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

is one possible solution.

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3 years ago