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tiny-mole [99]
3 years ago
13

A school typically sells 500 yearbooks in a year for $50 dollars each.

Mathematics
2 answers:
Evgen [1.6K]3 years ago
8 0
You divide 500 by $50 to get 10; the school sold 10 yearbooks that year if you’re looking for that as an answer.
HACTEHA [7]3 years ago
3 0

Answer:

Step-by-step explanation:

They should sell each yearbook 5 dollars cheaper because they will make 4,500 dollars more if they did that. With the discount and extra yearboos they will earn $29,250.

You might be interested in
What is the simplified form of the following expression 3 4x/5
Tanya [424]

Answer:

the correct answer is

3(4x/5)

therefore we multiple

12x/5

4 0
3 years ago
25- (3x + 5) = 2(x+8) + x how do you do it step by step?
NemiM [27]

Answer:

Simplifying

25 + -1(3x + 5) = 2(x + 8) + x

Reorder the terms:

25 + -1(5 + 3x) = 2(x + 8) + x

25 + (5 * -1 + 3x * -1) = 2(x + 8) + x

25 + (-5 + -3x) = 2(x + 8) + x

Combine like terms: 25 + -5 = 20

20 + -3x = 2(x + 8) + x

Reorder the terms:

20 + -3x = 2(8 + x) + x

20 + -3x = (8 * 2 + x * 2) + x

20 + -3x = (16 + 2x) + x

Combine like terms: 2x + x = 3x

20 + -3x = 16 + 3x

Solving

20 + -3x = 16 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

20 + -3x + -3x = 16 + 3x + -3x

Combine like terms: -3x + -3x = -6x

20 + -6x = 16 + 3x + -3x

Combine like terms: 3x + -3x = 0

20 + -6x = 16 + 0

20 + -6x = 16

Add '-20' to each side of the equation.

20 + -20 + -6x = 16 + -20

Combine like terms: 20 + -20 = 0

0 + -6x = 16 + -20

-6x = 16 + -20

Combine like terms: 16 + -20 = -4

-6x = -4

Divide each side by '-6'.

x = 0.6666666667

Simplifying

x = 0.6666666667

Step-by-step explanation:

7 0
3 years ago
What does p equal?pls help me.
Natalija [7]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the inequalities.

9p -9 <4p -2

===> 9p - 4p < -2 +9

===> 5p < 7

===> p < 7/5

hence the answer is P < 7/5

4 0
3 years ago
Please help !!!!!!!!!!!
ElenaW [278]

Answer:11/1 11/2 11/3 11/4 11/5  /6

Step-by-step explanation:

3 0
3 years ago
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
3 years ago
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