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3 years ago

Which system of equations does not have a real solution

Mathematics

1 answer:

Aloiza [94]3 years ago

5 0

Answer:

the correct answer is y = x 2 + 3x - 5 and x + y = -10

Step-by-step explanation:

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3.6399; thousandths round it

shepuryov [24]

Answer:

3.6399 = 3.640

Step-by-step explanation:

$3.6399\\1\:2\:3$

Rounding to nearest thousandth means rounding to 3 decimal place

(0 to 4) =round down

(5 to 9)= round up

3 0

3 years ago

Please help me with this again

dimaraw [331]

$( {x}^{2} -2x + 3)(x + 1) \\ {x}^{3} - 2 {x}^{2} + 3x + {x}^{2} - 2x + 3 \\ {x}^{3} - {x}^{2} + x + 3$

Therefore, the answer is the first choice.

3 0

3 years ago

Read 2 more answers

Help please!!!

katovenus [111]

Answer:

x = $\frac{7+\sqrt{47}\times i }{4}$

Step-by-step explanation:

<u>To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.</u>

x + 2y = 6 ---------------(1)

y - 5 = $(x-2)^{2}$ ---------------(2)

y = $(x-2)^{2}$ + 5 ---------------(3)

Substitute (3) in (1) ,

x + 2( $(x-2)^{2}$ + 5 ) = 6

$(a + b)^{2}$ = $a^{2} + 2ab + b^{2}$

x + 2( $x^{2} - 4x + 4$ + 5 ) = 6

$2x^{2} - 7x + 12=0$ --------------(4)

The roots of the quadratic equation $ax^{2} +bx+c$ is

x = $\frac{(-b) + \sqrt{(-b)^{2}-4 \times ac } }{2 \times a}$ -----------(5)

According to equation (5),solution of (4) is

x = $\frac{7 + \sqrt{(-7)^{2}-4 \times 24 } }{2 \times 2}$

x = $\frac{7+\sqrt{49-96}}{4}$

x = $\frac{7+\sqrt{47}\times i }{4}$

4 0

3 years ago

!!!KHAN ACADEMY!!! help please

barxatty [35]

Use photomath lollllll

3 0

2 years ago

What is the value of X and Y?

Marat540 [252]

Answer:

Y=20

X=10

Step-by-step explanation:

6 0

2 years ago