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Mariulka [41]
3 years ago
9

At which values of x does the graph of the function F(x) have a vertical asymptote? F(x)=x+2/x^2+2x-24 check all that apply.

Mathematics
1 answer:
GalinKa [24]3 years ago
8 0

Answer: X=4 X=-6

Step-by-step explanation:

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Order 34 × 102, 1.2 × 107, 8.11 × 10–3 and 435 from least to greatest.
andrey2020 [161]
I will assume that this question is about scientific notation (also assuming that the first number is 3.4*10^2), where you can arrange the numbers by the exponent of the 10, so from least to greatest:
8.11*10^{-3}, 3.4*10^{2}, 435, and 1.2*10^{7}
6 0
3 years ago
A) An empty bulldozer weighed 5 8 ⁄10 tons. If it scooped up 7 1 ⁄10 tons of dirt, what would be the combined weight of the bull
irakobra [83]

Answer: 12 9/10 and 10 7/8

Step-by-step explanation:

5 8/10 plus 7 1/10 equals 12 9/10

8 5/8 plus 2 2/8 equals 10 7/8

6 0
3 years ago
Can yall plz help me on this ???
Stolb23 [73]

Answer:

its the first one

Step-by-step explanation:

just use photomath (plus is free rn so you can see the steps)

4 0
3 years ago
Read 2 more answers
Please help me out with this equation! <br> i will mark you as brainliest
BigorU [14]
<h3>Answer:  -19, -15, -9, -1, 9   (choice A)</h3>

===================================================

Explanation:

If we plug in x = -2, then we get,

y = x^2 + 7x - 9

y = (-2)^2 + 7(-2) - 9

y = 4 - 14 - 9

y = -10 - 9

y = -19

So x = -2 leads to y = -19. The answer is between A and D.

---------

If you repeat those steps for x = -1, then you should get y = -15

Then x = 0 leads to y = -9

x = 1 leads to y = -1

Finally, x = 2 leads to y = 9

The outputs we get are: -19, -15, -9, -1, 9 which is choice A

Choice D is fairly close, but we won't have a second copy of -15, and  we don't have an output of -19.

7 0
3 years ago
Read 2 more answers
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
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