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mash [69]
3 years ago
14

Find the slope of a line through the points (0,0) and (2, 9). Round your

Mathematics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

Slope = 4.5

Step-by-step explanation:

We need to find the slope of a line through the points (0,0) and (2, 9).

We know that, the slope of a line is given by :

m=\dfrac{y_2-y_1}{x_2-x_1}

We have, x₁ = 0, x₂ = 2, y₁ = 0 and y₂ = 9

Put all the values,

m=\dfrac{9-0}{2-0}\\\\m=4.5

So, the slope of the line is 4.5.

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Let a=(1,2,3,4), b=(4,3,2,1) and c=(1,1,1,1) be vectors in R4. Part (a) [4 points]: Find (a⋅2c)b+||−3c||a. Part (b) [6 points]:
love history [14]

Solution :

Given :

a = (1, 2, 3, 4) ,    b = ( 4, 3, 2, 1),    c = (1, 1, 1, 1)     ∈   R^4

a). (a.2c)b + ||-3c||a

Now,

(a.2c) = (1, 2, 3, 4). 2 (1, 1, 1, 1)

         = (2 + 4 + 6 + 6)

         = 20

-3c = -3 (1, 1, 1, 1)

     = (-3, -3, -3, -3)

||-3c|| = $\sqrt{(-3)^2 + (-3)^2 + (-3)^2 + (-3)^2 }$

        $=\sqrt{9+9+9+9}$

       $=\sqrt{36}$

        = 6

Therefore,

(a.2c)b + ||-3c||a = (20)(4, 3, 2, 1) + 6(1, 2, 3, 4)  

                          = (80, 60, 40, 20) + (6, 12, 18, 24)

                         = (86, 72, 58, 44)

b). two vectors \vec A and \vec B are parallel to each other if they are scalar multiple of each other.

i.e., \vec A=r \vec B   for the same scalar r.

Given \vec p is parallel to \vec a, for the same scalar r, we have

$\vec p = r (1,2,3,4)$

$\vec p =  (r,2r,3r,4r)$   ......(1)

Let \vec q = (q_1,q_2,q_3,q_4)   ......(2)

Now given \vec p  and  \vec q are perpendicular vectors, that is dot product of \vec p  and  \vec q is zero.

$q_1r + 2q_2r + 3q_3r + 4q_4r = 0$

$q_1 + 2q_2 + 3q_3 + 4q_4  = 0$  .......(3)

Also given the sum of \vec p  and  \vec q is equal to \vec b. So

\vec p + \vec q = \vec b

$(r,2r,3r,4r) + (q_1+q_2+q_3+q_4)=(4, 3,2,1)$

∴ $q_1 = 4-r , \ q_2=3-2r, \ q_3 = 2-3r, \ q_4=1-4r$   ....(4)

Putting the values of q_1,q_2,q_3,q_4 in (3),we get

r=\frac{2}{3}

So putting this value of r in (4), we get

$\vec p =\left( \frac{2}{3}, \frac{4}{3}, 2, \frac{8}{3} \right)$

$\vec q =\left( \frac{10}{3}, \frac{5}{3}, 0, \frac{-5}{3} \right)$

These two vectors are perpendicular and satisfies the given condition.

c). Given terminal point is \vec a is (-1, 1, 2, -2)

We know that,

Position vector = terminal point - initial point

Initial point = terminal point - position point

                  = (-1, 1, 2, -2) - (1, 2, 3, 4)

                  = (-2, -1, -1, -6)

d). \vec b = (4,3,2,1)

Let us say a vector \vec d = (d_1, d_2,d_3,d_4)  is perpendicular to \vec b.

Then, \vec b.\vec d = 0

     $4d_1+3d_2+2d_3+d_4=0$

     $d_4=-4d_1-3d_2-2d_3$

There are infinitely many vectors which satisfies this condition.

Let us choose arbitrary $d_1=1, d_2=1, d_3=2$

Therefore, $d_4=-4(-1)-3(1)-2(2)$

                      = -3

The vector is (-1, 1, 2, -3) perpendicular to given \vec b.

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3 years ago
Step by step how to solve this equation by completing the square<br> x^2-16x+-60
anygoal [31]

Answer:

Step-by-step explanation:

x^2-16x+60

(x-6)(x-10) factor the equation

x-6=0

x=6

x-10=0

x=10

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3 years ago
the davis' backyard measures 40 ft x90 ft. find the area of their backyardand show how the area could be written using exponents
drek231 [11]
2,100 is the real number. 1,050² is exponent form.
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3 years ago
What is i^5????? Pls help
AnnZ [28]

Answer:

i^5 =

i

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Need help. i dont understand this!!!
Ilia_Sergeevich [38]

By the quadratic formula, the <em>solution</em> set of the <em>quadratic</em> equation is formed by two <em>real</em> roots: x₁ = 0 and x₂ = - 12.

<h3>How to find the solution of quadratic equation</h3>

Herein we have a <em>quadratic</em> equation of the form a · m² + b · m + c = 0, whose solution set can be determined by the <em>quadratic</em> formula:

x = - [b / (2 · a)] ± [1 / (2 · a)] · √(b² - 4 · a · c)      (1)

If we know that a = - 1, b = 12 and c = 0, then the solution set of the quadratic equation is:

x = - [12 / [2 · (- 1)]] ± [1 / [2 · (- 1)]] · √[12² - 4 · (- 1) · 0]

x = - 6 ± (1 / 2) · 12

x = - 6 ± 6

Then, by the quadratic formula, the <em>solution</em> set of the <em>quadratic</em> equation is formed by two <em>real</em> roots: x₁ = 0 and x₂ = - 12.

To learn more on quadratic equations: brainly.com/question/1863222

#SPJ1

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2 years ago
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