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postnew [5]
3 years ago
6

IS this corretct ? THANKYOU! :D WILL REWARD BRAINLIEST ANSWER!

Mathematics
1 answer:
AlladinOne [14]3 years ago
5 0
No, the answers are incorrect.
When reflected over the y-axis, if the x is positive, then it becomes negative, but if the x is negative, then it becomes positive. Also, the y will not change.
The answer should be 
X prime: (0,-6)
Y prime: (10,4)
Z prime: (-3,-1)
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If AD=20 and AC=3x+4, find the value of x. Then find AC and DC.<br><br> I don't understand.
Fofino [41]
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5 0
3 years ago
Any 5 <br>algebraic formulas
Ostrovityanka [42]

Answer:

a2 – b2 = (a – b)(a + b)

(a + b)2 = a2 + 2ab + b2

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4

a4 – b4 = (a – b)(a + b)(a2 + b2)

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6 0
2 years ago
Read 2 more answers
Polly stacked the tins from five boxes of cat food onto an empty shelf in a supermarket. There were 15 tins of cat food in each
Alexandra [31]

Answer:

(a) 3\frac{2}{3} boxes of cat food had been sold.

(b) \frac{11}{3} boxes of cat food had been sold.

Step-by-step explanation:

From the question,

Polly stacked the tins from five boxes of cat food onto an empty shelf and there were 15 tins of cat food in each box. That is,

5 × 15 tins of cat food were stacked onto the empty shelf.

5 × 15 = 75

Hence, 75 tins of cat food were stacked onto the empty shelf.

Also, from the question,

At the end of the week there were 20 tins left on the shelf and the rest of the tins had been sold, that is

75 - 20 tins had been sold

75 - 20 = 55

Hence, 55 tins of cat food had been sold.

To determine the number of boxes of cat food that had been sold,

Since there are 15 tins of cat food per box, then

we will divide 55 tins of cat food that had been sold by 15 tins of cat food per box

55 ÷ 15 = \frac{55}{15}

Hence, \frac{55}{15} boxes of cat food had been sold

(a) As a mixed number

\frac{55}{15}  = 3\frac{2}{3}

Hence, 3\frac{2}{3} boxes of cat food had been sold.

(b) As an improper fraction

\frac{55}{15}  = \frac{11}{3}

Hence, \frac{11}{3} boxes of cat food had been sold.

7 0
3 years ago
Anyone out here play fn or 2k? cause when i get my game bck im trynna run 10-15 only
Contact [7]

Answer:

I do but I only play fn not 2k

6 0
3 years ago
8
kolbaska11 [484]
No se la verdad!! Sube
8 0
2 years ago
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