For f(x) = 322 - x , find the value of f(-9).

Solve the system using submsitution x+y=-10 y=5x+2

Nitella [24]

Answer:

x+y= -10

y=5x+2

solution

x+(5x+2)= -10

x+5x+2= -10

6x+2= -10

6x= -10-2

6x= -12

6x/6= -12/6

x= -2

while y=5x+2

y= 5(-2)+2

y= -10+2

y= -8

x= -2, y= -8

4 0

3 years ago

Evaluate the following expression when x=-4 and y=4 x³-x /4y<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7By%7D" id="Tex

Simora [160]

Answer: - 60

Step-by-step explanation:

$\frac{(-4)^3 - (-4)}{4(4)^-4/4} \\\\= -60$

7 0

2 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

Solve the equation: 10 = -4 + x

Sonja [21]

10= -4 + x x=14 because you do the opposite of the sign you see in this case its a plus sign so you subtract -4 from ten and since the 4 is a negative and your subtracting you would add 4 I know this probably didn't help it sounds really confusing but I have hope that it did :)

8 0

3 years ago

Read 2 more answers

Items produced by a manufacturing process are supposed to weigh 90 grams. The manufacturing procon

Annette [7]

Using the normal distribution, it is found that 0.26% of the items will either weigh less than 87 grams or more than 93 grams.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 90 grams, hence $\mu = 90$ .

The standard deviation is of 1 gram, hence $\sigma = 1$ .

We want to find the probability of an item differing more than 3 grams from the mean, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3}{1}$

$Z = 3$

The probability is P(|Z| > 3), which is 2 multiplied by the p-value of Z = -3.

Looking at the z-table, Z = -3 has a p-value of 0.0013.

2 x 0.0013 = 0.0026

0.0026 x 100% = 0.26%

0.26% of the items will either weigh less than 87 grams or more than 93 grams.

For more on the normal distribution, you can check brainly.com/question/24663213

4 0

3 years ago