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Pani-rosa [81]
2 years ago
15

Describe the solutions of 5

Mathematics
1 answer:
djyliett [7]2 years ago
6 0

Answer:

A value, or values, we can put in place of a variable (such as x) that makes the equation true.

Example: x + 2 = 7

When we put 5 in place of x we get: 5 + 2 = 7

5 + 2 = 7 is true, so x = 5 is a solution

You might be interested in
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.

6 0
2 years ago
9) Three times a number added to 12 gives -6. Find the number.​
KatRina [158]

Answer:

The number is -6.

Step-by-step explanation:

Variable x = a number

Set up an equation:

3x + 12 = -6

Isolate variable x:

3x = -18

x = -6

Check your work:

3(-6) + 12 = -6

-18 + 12 = -6

-6 = -6

Correct!

7 0
3 years ago
Read 2 more answers
Pls help me with this problom asap!!!!
Nataly [62]

Answer:

4 1/2. Root 18 simplifies to 4.242640687, which is less than 4.50

8 0
2 years ago
Read 2 more answers
I really need help with this it’s on Aleks
liraira [26]

Answer:

y = 2

Step-by-step explanation:

Given

\sqrt[3]{3y+2} + 4 = 6 ( subtract 4 from both sides )

\sqrt[3]{3y+2} = 2 ( cube both sides )

3y + 2 = 2³ = 8 ( subtract 2 from both sides )

3y = 6 ( divide both sides by 3 )

y = 2

6 0
3 years ago
Read 2 more answers
Write three numbers grater than 20 that have 9 as a factor
Burka [1]
27, 36, 45 are greater than 20 and multiples of 9.
5 0
3 years ago
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