Help please!!!!!!!!!!!!!!!!!

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Over [174]

Answer: It's 64 with no remainder

8 0

3 years ago

Read 2 more answers

What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

Solution:

Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

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

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

$x=\frac{2 \pm \sqrt{-16}}{2}$

$x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}$

we know that square root of -1 is "i" which is a complex number

$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

6 0

3 years ago

How do you solve m - p/2 m=5 and p=2

navik [9.2K]

5-2/2
Divide first due to division coming before addition in pemdas
5-1
4

7 0

3 years ago

Please tell the answer and show the work.

LUCKY_DIMON [66]

First step, i would try to figure out the slope. think of it as a typical y=mx+b linear equation as you check out the graph.

it's just rise over run (rise up this many, run over that many), and your options are (-2/3) or (-3/2)

start at the y-intercept, which is just above the middle of the graph. if you go down 2 and try to go over 3, you hit the line before you can count 3 units. that means this slope is incorrect.

go back to the y-intercept. down 3, over 2--there you go. your slope here is (-3/2), which immediately gets rid of half your answer choices for having the wrong slope

the next thing you have to do is decide where this graph should be shaded. the shaded region shows the domain; the white region shows what's outside of your domain. this inequality is shaded "below" which means that the y values are LESS than "(-3/2)x +1"

that eliminates another one of your answer choices; choice 3 is correct. the inequality is y < (-3/2)x + 1

if the graphs of inequalities are shaded below and the line is dotted, it's less than. if the graphs of inequalities are shaded above and the line is dotted, it's greater than. general rules for ya

5 0

3 years ago

Read 2 more answers

Match the systems of linear equations with their solutions.

OleMash [197]

Answer:

The solutions of linear equations in the procedure

Step-by-step explanation:

Part 1) we have

x+y=-1 ----> equation A

-6x+2y=14 ----> equation B

Solve the system by elimination

Multiply the equation A by 6 both sides

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

6x+6y=-6 -----> equation C

Adds equation C and equation B

6x+6y=-6

-6x+2y=14

-------------------

6y+2y=-6+14

8y=8

y=1

Find the value of x

substitute in the equation A

x+y=-1 ------> x+1=-1 ------> x=-2

The solution is the point (-2,1)

Part 2) we have

-4x+y=-9 -----> equation A

5x+2y=3 ------> equation B

Solve the system by elimination

Multiply the equation A by -2 both sides

-2*(-4x+y)=-9*(-2)

8x-2y=18 ------> equation C

Adds equation B and equation C

5x+2y=3

8x-2y=18

----------------

5x+8x=3+18

13x=21

x=21/13

Find the value of y

substitute in the equation A

-4x+y=-9 ------> -4(21/13)+y=-9 ----> y=-9+84/13 -----> y=-33/13

The solution is the point (21/13,-33/13)

Part 3) we have

-x+2y=4 ------> equation A

-3x+6y=11 -----> equation B

Multiply the equation A by 3 both sides

3*(-x+2y)=4*3 ------> -3x+6y=12

so

Line A and Line B are parallel lines with different y-intercept

therefore

The system has no solution

Part 4) we have

x-2y=-5 -----> equation A

5x+3y=27 ----> equation B

Solve the system by elimination

Multiply the equation A by -5 both sides

-5*(x-2y)=-5*(-5)

-5x+10y=25 -----> equation C

Adds equation B and equation C

5x+3y=27

-5x+10y=25

-------------------

3y+10y=27+25

13y=52

y=4

Find the value of x

Substitute in the equation A

x-2y=-5 -----> x-2(4)=-5 -----> x=-5+8 ------> x=3

The solution is the point (3,4)

Part 5) we have

6x+3y=-6 ------> equation A

2x+y=-2 ------> equation B

Multiply the equation B by 3 both sides

3*(2x+y)=-2*3

6x+3y=6

so

Line A and Line B is the same line

therefore

The system has infinite solutions

Part 6) we have

-7x+y=1 ------> equation A

14x-7y=28 -----> equation B

Solve the system by elimination

Multiply the equation A by 7 both sides

7*(-7x+y)=1*7

-49x+7y=7 -----> equation C

Adds equation B and equation C

14x-7y=28

-49x+7y=7

------------------

14x-49x=28+7

-35x=35

x=-1

Find the value of y

substitute in the equation A

-7x+y=1 -----> -7(-1)+y=1 ----> y=1-7 ----> y=-6

The solution is the point (-1,-6)

6 0

3 years ago