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a_sh-v [17]
3 years ago
12

Order the numbers from least to greatest. -7/5, -1 5/11, -1.45, -1 91/200​

Mathematics
1 answer:
Anna35 [415]3 years ago
7 0

Answer:

Step-by-step explanation:

-7/5=-1.400

-1 5/11=-16/11=-1.455...

-1.450

-1  91/200=-291/200=-1.405

-1.455 <-1.450 <-1.405 <-1.400

- 1 5/11 <-1.45 <- 91/200 <-7/5

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The ratio of marbles to stones is 5 to 7 if there are 35 marbles how many stones is there?
icang [17]
5 : 7
Marbles : Stones

35/5 = 7
So stones
= 7 * 7
49

=35 : 49
3 0
3 years ago
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The number p and −1.2 are additive inverses. Drag and drop −1.2 and p to their correct positions on the number line. Drag and dr
klemol [59]

Answer:

p=1.2

Step-by-step explanation:

Additive inverse of any number means a number which can be added to the original number to get

Two numbers are said to be additive inverse of each other if sum of both numbers is 0. For example if 5 is the given number then -5 will be it's additive inverse. So to find additive inverse, we basically change the sign of number.

Given number p is additive inverse off -1.2 so p must be 1.2.

Sum will obviously be 0.

So to graph them on number line, we make a point at -1.2, 1.2 and at 0 for their sum.


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3 years ago
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Compare the pair of fractions. ( &lt;,&gt;,=) <br> 3/4__5/6
Andrew [12]

Answer:

3/4 is less < than 5/6

Step-by-step explanation:

4 0
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.  

5 0
3 years ago
Which table identifies the one-sided and two-sided limits of function at x = 2?​
Yuliya22 [10]

Answer:

Table 3

Step-by-step explanation:

Check table three;

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lim\:_{x\to \:2^+}\:f\left(x\right)=\:1

Since the left hand limit (lim\:_{x\to \:2^-}\:f\left(x\right)) is not equal to the right hand limit (lim\:_{x\to \:2^+}\:f\left(x\right)),  the limit as x approaches to 2 does not exist.

Therefore "nonexistent" is true, and table 3 is the correct model of the limits of the function at x = 2

6 0
3 years ago
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