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ehidna [41]
2 years ago
15

A) 3√2 - 5√8 + 2√20

Mathematics
1 answer:
tigry1 [53]2 years ago
7 0

Answer:

,bn ljb ib

Step-by-step explanation:

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Write x^2 + 6x -7 in the form (x+a)^2 +b where a and b are integers please
ANTONII [103]

Answer:

(x + 3) {}^{2}  - 16

Step-by-step explanation:

{x}^{2}  + 6x - 7 = (x  + 3)  {}^{2}  - (3) {}^{2}  - 7 \\  = (x + 3) {}^{2}  - 16

3 0
3 years ago
Why is the product of 9s diagonal?
Nadusha1986 [10]

Answer:

A. With each 9s fact you add another 9, which means you go straight down a row and back one space. This creates a diagonal pattern in the grid.

6 0
3 years ago
Question 2<br> Solve by substitution.<br> 2x – 10y = 14<br> 2x + 6y = -2
Fed [463]

Answer:

x= 2 and y= -1

Step-by-step explanation:

x=7 +5y

14+16y=−2

y=−1

Substitute y=-1y=−1 into x=7+5yx=7+5y.

x=2

8 0
3 years ago
X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
Please give the answer
dybincka [34]

Answer:

When N is an even number, use the first equation in the matrix. When N is negative use the second equation.

Step-by-step explanation:

n = 6 | 6/2 = 3

n=5 | 3(5)+1 = 16

5 0
2 years ago
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