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AleksandrR [38]
3 years ago
7

Which of these points lie on the line described by the equation below?

Mathematics
2 answers:
nataly862011 [7]3 years ago
6 0

Answer:

<u>Answer</u><u>:</u><u> </u><u>A</u><u>.</u><u> </u><u>(</u><u>6</u><u>,</u><u> </u><u>4</u><u>)</u>

step by step

Alex777 [14]3 years ago
3 0

Answer:

<u>Answer</u><u>:</u><u> </u><u>A</u><u>.</u><u> </u><u>(</u><u>6</u><u>,</u><u> </u><u>4</u><u>)</u>

Equation of the line:

y - 4 =  - 2(x - 6) \\ y - 4 =  - 2x + 12 \\ { \underline{y =  - 2x + 16}}

when x is 6:

y =  - 2(6) + 16 \\ y =  - 12 + 16 \\ y = 4

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Which of the following are NOT sufficient to prove that a quadrilateral is a parallelogram?
Ksenya-84 [330]
The answers are
<span>I. Two pairs of opposite angles congruent. 
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3 0
3 years ago
Read 2 more answers
In what time will Rs. 5600 amount to Rs. 6720 at 8% per annum?​
Leona [35]

Answer: 2 years

Step-by-step explanation:

If Interest= Principal x rate x time/100; then time is (100 × Interest)/(Principal × Rate)

From the question, old Principal is 5600, new principal =6720, the difference is the interest accrued.

Interest = 6720 - 5600 = Rs 1120

Rate = 8% = 8 ÷ 100 = 0.08

Time = x

Then, slot the values into the formula

Time= 100 x 1120 / 6720 x 8

= 112000/53760

=2.08

Time= 2years

I hope this helps

8 0
4 years ago
Really really need help on ONE math graph question.
Nat2105 [25]
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7 0
3 years ago
What is the ​ IQR ​ for the data set? {31, 30, 40, 35, 48, 44, 25} Drag and drop the correct answer into the box.
Kitty [74]

Answer:

B

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4 0
2 years ago
The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:
deff fn [24]

Step-by-step explanation:

Given equation of circle is,

{x}^{2}  +  {y}^{2}  - 6y - 12 = 0 \\ by \: compairing \: it \: with \:   \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f=  - 6 \\ or \: f=  - 3 \\ c =  - 12

again the another form of circle is,

{x}^{2}  +  {(y - k)}^{2}  = 21 \\ or \:  {x}^{2}  + {y}^{2}  - 2ky +  {k}^{2}  - 21 = 0 \\ by \: compairing \:it \: with \:  \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get  \\ g = 0 \\f =  - k \\ c =  {k}^{2}  - 21

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3

4 0
3 years ago
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