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drek231 [11]
3 years ago
13

X - 2y + 3z= -9 2x + 5y + z = 10 3x – 6y + 9z = 12

Mathematics
1 answer:
Kobotan [32]3 years ago
5 0

solve\:for\:x,\:x-2y+3z=-9\quad :\quad x=-9+2y-3z

solve\:for\:x,\:2x+5y+z=10\quad :\quad x=\frac{10-5y-z}{2}

solve\:for\:x,\:3x-6y+9z=12\quad :\quad x=4+2y-3z

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Ronald measured a house and its lot and made a scale drawing. The front patio, which is 14 meters wide in real life, is 7 millim
zhannawk [14.2K]

Answer:

2 metre= 1millimetre

Step-by-step explanation:

From the measurement of the house The front patio, is 14 meters wide in real life, is 7 millimeters wide in the drawing

Real life= 14 meters

Drawing= 7 millimeters,

The scale he used is 2 metre= 1millimetre

Which means he magnify the drawing by scale of 2 metre to get the real life structure.

✓Let us clarify it

14 meters = 7 millimeters

2 metre= x millimeter

If we cross multiply

7 millimeters× 2 metre= 14 metre × X millimeters

Make X subject of the formula we have

X= 1millimetres

3 0
3 years ago
I need help with this !! chart
Burka [1]

Answer:

Make coordinates with this table.

Step-by-step explanation:

What I think from these tables is that you can make coordinates with these points. Here are the instructions for you to follow-

  1. See the "y" and "x". These are the axis. So make coordinates with these. Ex- (0,3), (1,5) and so on
  2. Plot these coordinates in a graph paper.
  3. After plotting them, you might find the shape it made.

I hope it helps you.

3 0
3 years ago
What number is needed to complete the Pattern next 59, 51, 44, 38, 33, 29, 26, next
Vedmedyk [2.9K]

Answer: 22


Step-by-step explanation:

To get from 59 to 51 you subtract 8 then it was minus 7, then minus 6, then minus 5, and then it just kept going down like that

4 0
3 years ago
Read 2 more answers
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
Which of the answers are correct? Select all that apply.
Leona [35]

Answer:A, b and also c

Step-by-step explanation:

8 0
3 years ago
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