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NISA [10]
4 years ago
6

Help!!! Will give 10 points!

Mathematics
1 answer:
k0ka [10]4 years ago
8 0

Answer:

1/1 which converts to a slope of 1

Step-by-step explanation:

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Order from least to greatest. -1/5,1/7,-0.5,1/8
Ivenika [448]

There ya go, Hope, this helps

5 0
4 years ago
I need the answer plz I’ll give brain
Aleksandr [31]
The answer to this problem would be A, $1.15. You get this answer by taking the total cost and dividing that by how many yards of fabric, for example let’s take the first one. $2.53 divided by 2.2 would be $1.15, same goes for the rest.
3 0
4 years ago
I need help arranging cards in a 4 x 4 grid with numbers 1-9 so that the sum of each row and each column is 20. This what the gr
igor_vitrenko [27]

Answer:

8, 2, 8, 2

7, 3, 7, 3

6, 4 6, 4

9, 1, 9, 1

Step-by-step explanation:

In the picture, it shows two of each number so I'm gonna base it off the picture but let me know if I should base it off the wording.

So we know that half of 20 is 10. This will be valuable information. So we need the sum of two numbers to equal 10. So for the rows we can just do 8+2+8+2, 7+3+7+3, 6+4+6+4, 9+1+9+1.

Since there are 18 numbers, I'm assuming we don't have to use all of them.

3 0
3 years ago
Read 2 more answers
Could I have help if any of these problems?
valentinak56 [21]
Question 27 The answer is x = 1 4/11
7 0
4 years ago
MATRECIES: Suppose ={u,v,w}⊆R3 is linearly independent. Determine if the set 4 ={u, v, v−w, u+v+w} is linearly independent or li
AlladinOne [14]

The set is linearly dependent.

To explicitly prove this, we need to show there is at least one choice of constants c_1,c_2,c_3,c_3\in\Bbb R such that

c_1 u + c_2 v + c_3(v-w) + c_4(u+v+w) = 0

or equivalently,

(c_1 + c_4) u + (c_2 + c_3 + c_4) v + (-c_3 + c_4) w = 0

which is the same as solving the system of equations

\begin{cases} c_1 + c_4 = 0 \\ c_2 + c_3 + c_4 = 0 \\ -c_3 + c_4 = 0 \end{cases}

From the first and last equations, we have c_1=-c_4 and c_3=c_4. Substituting these into the second equation leaves us with c_2-2c_1=0, and so the overall solution set is

c_1 = \dfrac12 c_2 = -c_3 = -c_4

for which there are infinitely many not-all-zero solutions.

3 0
2 years ago
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