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melamori03 [73]
3 years ago
6

The percent markup is ____​%

Mathematics
1 answer:
morpeh [17]2 years ago
4 0

Answer:

the percent is almost 42.9

Step-by-step explanation:

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Brynne has 13 books she has 8 more books than games how many game does Brynne have
vagabundo [1.1K]

Answer:

5

Step-by-step explanation:

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3 years ago
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Richard’s checking account balance was $57.34 at the beginning of the week. During the week, he recorded the transactions below.
Tomtit [17]

Answer:

US$ 132.45

Step-by-step explanation:

See attachment for the missing table.

Given:

Richard’s checking account balance at the beginning of the week = $57.34

<u>Richard’s account balance at the end of the week from the given table:</u>

Deposits of the week = US$ 163.75

Expenses of the week = Groceries + Credit card bill + Gas

Expenses of the week = 25.37 + 50 + 13.27

Expenses of the week = US$ 88.64

Richard’s account balance at the end of the week = Richard’s checking account balance at the beginning of the week + Deposits of the week - Expenses of the week

Replacing with the real values:

Richard’s account balance at the end of the week = 57.34 + 163.75 - 88.64

                                                                                    =US$ 132.4

7 0
3 years ago
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Refer to the figure below to complete the following problem.
IRINA_888 [86]

Answer:

38

Step-by-step explanation:

5x-7=3x+ll

5x=3x+18

2x/2=18/2

x    =9

5 x 9 -7 = 36

3 x 9 + 11=36

m D =36

3 0
3 years ago
All vectors are in Rn. Check the true statements below:
Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that ||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then ||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||.

C) Consider S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2. This set is orthogonal because (0,2)\cdot(2,0)=0(2)+2(0)=0, but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in \mathbb{R}^n. Then the columns of A form an orthonormal set. We have that A^{-1}=A^t. To see this, note than the component b_{ij} of the product A^t A is the dot product of the i-th row of A^t and the jth row of A. But the i-th row of A^t is equal to the i-th column of A. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then A^t A=I    

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set \{u_1,u_2\cdots u_p\} and suppose that there are coefficients a_i such that a_1u_1+a_2u_2\cdots a_nu_n=0. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then a_i||u_i||=0 then a_i=0.  

5 0
3 years ago
Add the two expressions. 4.6x−3 and −5.3x+9 Enter your answer, in simplified form, in the box. PLEASE ANSWER FAST THIS IS A BIG
Ivan
To add the two expressions, we can write it as:

4.6x-3 + (-5.3x+9)

We can distribute the plus sign (which means just drop the parenthesis in this case):

4.6x-3-5.3x+9

Now, we can simplify by combining like terms:

-0.7x+6
4 0
3 years ago
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