Yo can someone pass a brainly plus account

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17 eq(1)

y = -1/2x eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 5/4, b =0 and c = -17

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

System A has 4 real solutions.

System B

y = x^2 -7x + 10 eq(1)

y = -6x + 5 eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-1 and c =5

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9 eq(1)

8x - y = -17 eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17

8(-2) - y = -17

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0

3 years ago

Read 2 more answers

Ax+by+c=0 solve for y

kvv77 [185]

$Answer: \large\boxed{y=\frac{-ax}{b} -\frac{c}{b}}$

Step-by-step explanation:

$ax+by+c=0\\ax+by=-c\\by=-ax-c\\y=\frac{-ax-c}{b}\\y=\frac{-ax}{b} -\frac{c}{b}$

8 0

2 years ago

(8,6) for 2 + y = x and -2y + 20 = x <br> Can someone help me please.

ehidna [41]

Answer:

3939π

Step-by-step explanation:

The new system is not the only one who thinks that he was a member of the American to the right of way of the most important things in life is to make sure the first half of an old friend who was in the the ↕️

5 0

2 years ago

Help please. I need to finish this in five minutes!

e-lub [12.9K]

Answer:

i have no idea....

Step-by-step explanation:

but i hope someone helps :(( good luck:((

3 0

3 years ago

What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If