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2 years ago

Please help 11 points worth

Mathematics

2 answers:

DochEvi [55]2 years ago

6 0

Answer:

positive

cuz i just know :)

good luck

navik [9.2K]2 years ago

5 0

It’s definitely Positive

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How much money should be deposited today in an account that earns 7% compounded semiannually so that it will accumulate to $12,0

Korvikt [17]

Given rate is = 7% or 0.07

Total amount needed = $12000

Time = 4 years

Here, the deposit is compounded semiannually, means twice per year and this gives 8 annual compounding periods in 4 years.

The equation becomes:

P= $\frac{12000}{(1+\frac{0.07}{2})^{2.4}}$

P = $\frac{12000}{1.035^{8}}$

Solving it, we get P = $ 9112.93

Hence $9112.93 should be deposited today.

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3 years ago

Please help me with these questions! Thanks

sergiy2304 [10]

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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST Which of the following scale factors will result in an enlargement? A. k < –1 B. –1 < k &

Sergio [31]

Answer: OPTION A.

Step-by-step explanation:

By definition, a dilation can be an enlargement or a reduction of the shape.

An enlargement is when dilation creates a larger image and a reduction is when dilation creates a smaller image.

When, the scale factor is greater than 1, the image is an enlargement and when the scale factor is between 0 and 1, the image is a reduction.

It is important to know that with a negative scale factor the enlargement will will be inverted and it will also be on the other side of the center of dilation.

Knowing this, we can say that the scale factor that will result in an enlargement is:

$k$

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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 $a_i=0$ .