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3 years ago

What is the opposite of -4

Mathematics

2 answers:

Marina86 [1]3 years ago

5 0

The answer is 4!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Katen [24]3 years ago

3 0

4 is the opposite of -4.

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Write the equation of the line perpendicular to 2x+3y=9 that passes through (-2,5). Write your answer in slope-intercept form. S

Harman [31]

ANSWER

$y = \frac{3}{2} x + 8$

EXPLANATION

The line given to us has equation,

$2x + 3y = 9$

We need to write this equation in the slope intercept form to obtain,

$3y = - 2x + 9$

$\Rightarrow \: y = - \frac{2}{3}x + 3$

The slope of this line is

$m_1 = - \frac{2}{3}$
Let the slope of the perpendicular line be

$m_2$

Then
$m_1 \times m_2 = - 1$

$- \frac{2}{3} m_2= - 1$

This implies that,

$m_2 = - 1 \times - \frac{3}{2}$

$m_2 = \frac{3}{2}$

Let the equation of the perpendicular line be,

$y = mx + b$

We substitute the slope to get,

$y = \frac{3}{2} x + b$

Since this line passes through
$(-2,5)$
it must satisfy its equation.

This means that,

$5= \frac{3}{2} ( - 2)+ b$

$5 = - 3 + b$

$5 + 3 = b$

$b = 8$

Wherefore the slope-intercept form is

$y = \frac{3}{2} x + 8$

6 0

3 years ago

The image of (4, 1) under the transformation T : (x, y) → (x + 3, y + 1) is

larisa [96]

(4 + 3 , 1 + 1)
(7,2)

4 0

3 years ago

From the hay loft door, Ted sees his dog on the ground. The angle of depression of the dog is 40º. Ted's eye level is 16 feet ab

Vlad1618 [11]

Answer: 19 feet

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the next trigonometric function.

Tan α = opposite side / adjacent side

Where α is the angle of depression of the dog, the opposite side (16) is Ted's eye level above the ground, and the adjacent side (x) is the distance between the dog and the barn door.

Replacing with the values given:

tan40 = 16/x

Solving for x

x =16/tan40

x= 19 ft

Feel free to ask for more if needed or if you did not understand something.

5 0

3 years ago

The diagram shows a cone and its axis of rotation. Which type of cross section is formed when the cone is intersected by a plane

yKpoI14uk [10]

Answer:

Right-triangle

Step-by-step explanation:

Take a right triangle for example, then twist it 360 degrees, keeping the longest leg at the center of rotation. It will then form a cone.

8 0

3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago