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.

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>

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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.

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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