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3 years ago

HELP IMMEDIATELY!!

Mathematics

2 answers:

marta [7]3 years ago

8 0

Answer:

- 3

(got it right on edmentum)

<h2>but for your assurance so you can get this question right please make sure to look at the picture below↓</h2>

Rashid [163]3 years ago

6 0

Answer:

The value of c is 3

Step-by-step explanation:

Here, we want to get the value of c

From the question, we can select a single x-point

when x = 16

f(x) = 2

g(x) = -1

Thus;

g(16) = f(16) - c

-1 = 2 - c

c = 2 + 1

c = 3

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Describe the transformations of how the graph y=x^2 can be transformed into y= -1/5 (x-4)^2 + 2

Orlov [11]

Answer:

Reflected over the x-axis, vertically shrunk by a factor of 1/5, moved right 4 units and up 2 units.

Step-by-step explanation:

Parent Graph: f(x) = a(bx - h)² + k

a is vertical stretch (a > 1) or shrink (a < 1) and reflection over x-axis

b is horizontal stretch or shrink and reflection over y-axis

h is horizontal movement left (if positive) or right (if negative)

k is vertical movement up (if positive) or down (if negative)

Now that we have our rules, we simply apply it to y = -1/5(x - 4)² + 2

a = -1/5, so reflection over x-axis and vertical shrink of 1/5 (1/5 < 1)

Nothing has changed b, so no horizontal stretch/shrink

h = -4, so horizontal movement right 4 units

k = 2, so vertical movement up 2 units

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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

A local company manufactures netbook computers. Their profit function is given by this equation: y equal minus x to the power of

Westkost [7]

Answer by BlueSky06

The equation described above can also be written as, y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero. dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit. y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day.
Read more on Brainly.com - brainly.com/question/3586459#readmore

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2 years ago

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bonufazy [111]

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3 years ago

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