1.64 points as a mix number in simplest form

Denise makes three times as many chocolate cupcakes as vanilla cupcakes. She also makes three dozen strawberry cupcakes. If Deni

slamgirl [31]

Answer:

6 dozens

Step-by-step explanation:

Vanilla, v = x

Chocolate, c = 3x

Strawberry, s = 3 dozens = 36

Total cupcakes = 132

Hence,

x + 3x + 36 = 132

4x = 96

x = 96 / 4

x = 24

Chocolate = 3 * 24 = 72

Dozens of chocolate = 72 /12 = 6 dozens

7 0

3 years ago

What is the slope of the line shown in the graph? (4 points)

Korvikt [17]

Answer:

a. or $\frac{3}{-2}$

Step-by-step explanation:

Slope is equal to y1-y2 over x1-x2.

So the equation would be:

$\frac{2-(-1)}{-3-(-1)}$

2-(-1)=2+1=3

-3-(-1)=-3+1=-2

So the answer is:

$\frac{3}{-2}$

5 0

3 years ago

Nicola buys a bouquet of 8 sunflowers for $18 What is the cost of 1 sunflower?

Sever21 [200]

Answer:

$2.25

Step-by-step explanation:

Divide the cost ($18) by the amount of flowers (8.)

7 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 $a_i=0$ .