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2 years ago

I need to know how to work it out someone plz help me out due tommorow

Mathematics

1 answer:

DiKsa [7]2 years ago

5 0

Multiply 75 by 1.0675 to find the total amount of tax plus 100 percent of the original fee im pretty sure and for the seocnd one multiply 450 by 1.45

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Given the arithmetic sequence an = 2 + 4(n − 1), what is the domain for n?

zlopas [31]

The answer should be A the first one

3 0

2 years ago

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At 6:00 A.M. the temperature was 33°F. By noon the temperature had increased by 10°F and by 3:00 P.M. it had increased another 1

chubhunter [2.5K]

So let's see here...

6:00 A.M. temperature is 33°F.

12:00 P.M temperature increased by 10°F making it 43°F.

3:00 P.M. temperature increased by another 12°F making it 55°F.

At 10:00 P.M it would decrease by 15°F making it 40°F.

The temperature would need to fall (or decrease) by 7°F to reach the original temperature of 33°F

Hope this helped!

3 0

2 years ago

On Triangle ABC, Angle A measures 30 degrees, Angle B measures 90 degrees, and Angle C measures 60 degrees. List the sides of Tr

Mazyrski [523]

Answer:

A=30 B=90 C=60

Step-by-step explanation:

A=30 C=60 B=90

8 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 $a_i=0$ .