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2 years ago

9

Suppose the function S ( t ) = 2.5 cos ( 0.017 t − 1.35 ) + 7 models the amount of snowfall for a year in Anchorage, Alaska, whe

re t = 1 is the first day of the year. Based on the function, what is the range of snowfall for Anchorage?

1 answer:

BlackZzzverrR [31]2 years ago

8 0

Answer:

4.5 to 9.5

Step-by-step explanation:

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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

Square root of 110 - n

AlexFokin [52]

Answer:

√(110 - n)

Step-by-step explanation:

This can't be simplified. All we can do here is re-write "square root of 110 - n" symbolically, which comes out to √(110 - n). The parentheses are essential for clarity.

8 0

2 years ago

What is the simplified base of the function f(x) = One-fourth (Root Index 3 StartRoot 108 EndRoot) Superscript x?

Rainbow [258]

Answer:

The base is: $3 \sqrt[3]{4}$

Step-by-step explanation:

Given

$f(x) = \frac{1}{4}(\sqrt[3]{108})^x$

Required

The base

Expand 108

$f(x) = \frac{1}{4}(\sqrt[3]{3^3 * 4})^x$

Rewrite the exponent as:

$f(x) = \frac{1}{4}(3^3 * 4)^\frac{1}{3}^x$

Expand

$f(x) = \frac{1}{4}(3^3^\frac{1}{3} * 4^\frac{1}{3})^x$

$f(x) = \frac{1}{4}(3 * 4^\frac{1}{3})^x$

Rewrite as:

$f(x) = \frac{1}{4}(3 \sqrt[3]{4})^x$

An exponential function has the following form:

$f(x)=ab^x$

Where

$b \to base$

By comparison:

$b =3 \sqrt[3]{4}$

So, the base is: $3 \sqrt[3]{4}$

4 0

2 years ago

A local kids play sold 8 adult tickets and 9 child tickets for a total of $205. On day two they sold only 4 adult tickets and 3

Naya [18.7K]

Answer:

The price of 1 adult ticket is $ 11 and price of 1 child ticket is $13

Step-by-step explanation:

Let x be the cost of 1 adult ticket

Let y be the cost of 1 child ticket

Cost of 8 adult tickets = 8x

Cost of 9 child tickets = 9y

We are given that A local kids play sold 8 adult tickets and 9 child tickets for a total of $205

So, 8x+9y=205 ----- 1

Cost of 4 adult tickets = 4x

Cost of 3 child tickets = 3y

We are given that 4 adult tickets and 3 child tickets for a total of $83.

4x+3y=83 ----2

Plot 1 and 2 on graph

8x+9y=205 -- Red line

4x+3y=83 -- Blue line

Intersection point provides the solution

So, Intersection point =(11,13)

So, The price of 1 adult ticket is $ 11 and price of 1 child ticket is $13

4 0

2 years ago

What is ghe solution to this equation?<br> 02(x - 20) = 44 - x

Oliga [24]

X= 40 is the answer to this equation

5 0

2 years ago