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2 years ago

The exponential function g, whose graph is given below, can be written as g(x)=a*b^x

Mathematics

1 answer:

sveticcg [70]2 years ago

7 0

Answer:

g(x) = $9(\frac{1}{3})^x$

Step-by-step explanation:

Exponential function 'g' that can be written as g(x) = a.bˣ

From the graph attached,

Graph of the function passes through two points (0, 9) and (1, 3)

g(0) = a.b⁰

9 = a

Similarly, g(1) = a.b¹

3 = a.b

3 = 9(b)

b = $\frac{3}{9}=\frac{1}{3}$

Therefore function will be,

g(x) = $9(\frac{1}{3})^x$

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Answer:

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Step-by-step explanation:

Because of this fact:

726 > 660 and 726 < 749

So his credit rating is considered as Good ( 660- 749)

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Rina8888 [55]

Answer:

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Step-by-step explanation:

one batch calls for:

3/4 PB

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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 $a_i=0$ .

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3 years ago

Roy Gross is considering an investment that pays 7.6 percent, compounded annually. How much will he have to invest today so that

Debora [2.8K]

Answer: he should invest $16129 today.

Step-by-step explanation:

Let $P represent the initial amount that should be invested today. It means that principal,

P = $P

It would be compounded annually. This means that it would be compounded once in a year. So

n = 1

The rate at which the principal would be compounded is 7.6%. So

r = 7.6/100 = 0.076

The duration of the investment would be 6 years. So

t = 6

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years.

A = 25000

Therefore

25000 = P(1+0.076/1)^1×6

25000 = P(1.076)^6

25000 = 1.55P

P = 25000/1.55

P = $16129

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3 years ago