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3 years ago

Find the slope of the line containing the two points. (0,3),(4,0) Please show work

Mathematics

1 answer:

Delvig [45]3 years ago

5 0

Answer:

$m=-\frac{3}{4}$ .

Step-by-step explanation:

We have been given coordinates of two points on a line. We are asked to find the slope of the line using given points.

We will use slope formula to solve our given problem.

$m=\frac{y_2-y_1}{x_2-x_1}$ , where,

m = Slope of line,

$y_2-y_1$ = Difference between two y-coordinates on the line.

$x_2-x_1$ = Difference between two x-coordinates of the same y-coordinates.

Let $(0,3)=(x_1,y_1)$ and $(4,0)=(x_2,y_2)$

Substitute coordinates of the given points in slope formula:

$m=\frac{0-3}{4-0}$

$m=\frac{-3}{4}$

$m=-\frac{3}{4}$

Therefore, the slope of the line passing through given points is $-\frac{3}{4}$ .

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Answer:

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Step-by-step explanation:

1/6z=-10

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3 years ago

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3 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much

Y_Kistochka [10]

Answer:

x = ∛ 2*V/5

y = ∛ 2*V/5

h = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h where h = V / x*y

As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h

According to this:

Ct (cost of aquarium ) = cost of the base + cost of the sides

cₐ ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h

C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or

C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(y) = 5*c₁*x - 2*c₁*V/y²

C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²

or 5*y - 2*V/x² = 5*x - 2*V/y²

(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²

5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)

by symmetry x = y

Then using x = y and plugging that value on the derivatives

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(x) = 5*c₁*x - 2*c₁*V/x²

C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0

5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5

x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25

7 0

3 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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1 year ago

9i-2 = 6i+ 19 distributive property

Snezhnost [94]

Answer:

i=7

Step-by-step explanation:

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3 years ago