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3 years ago

HELP ASAP!!!! Please!!!! :) really need this.

Mathematics

1 answer:

Gnom [1K]3 years ago

8 0

Hello fellow k12 student um i would think it is the second one :)

Hope this helped :)

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Hey i need help thank you!

chubhunter [2.5K]

Answer:

I guess its like this...

not sure if its ryt! :(

3 0

3 years ago

the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.

Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

$2x^2-3x-7+7=0+7$

$2x^2-3x=7$

Divide both sides by 2

$\frac{2x^2-3x}{2}=\frac{7}{2}$

$x^2-\frac{3x}{2}=\frac{7}{2}$

Add (-3/4)² to both sides

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{65}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{3}{4}=\sqrt{\frac{65}{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}$

$x-\frac{3}{4}=\frac{\sqrt{65}}{4}$

Add 3/4 to both sides

$x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}$

$x=\frac{\sqrt{65}+3}{4}$

similarly solving

$x-\frac{3}{4}=-\sqrt{\frac{65}{16}}$

$x=\frac{-\sqrt{65}+3}{4}$

So the solution of the equation will have the values of x such as:

$x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}$

6 0

3 years ago

Which of the following is the graph of the line y=-3/4x+5 ?

Naddika [18.5K]

Slope : -3/4
Y -intercept (0,5)

3 0

3 years ago

M∠LON=77 ∘ m, angle, L, O, N, equals, 77, degrees \qquad m \angle LOM = 9x + 44^\circm∠LOM=9x+44 ∘ m, angle, L, O, M, equals, 9,

timama [110]

Answer:

Step-by-step explanation:

Given

<LON = 77°

<LOM = (9x+44)°

<MON = (6x+3)°

The addition postulate is true for the given angles since tey have a common point O:

<LON = <LOM+<MON

Since we are not told what to find we can as well look for the value of x, <LOM and <MON

Substitute the given parameters and get x

77 = 9x+44+6x+3

77 = 15x+47

77-47 = 15x

30 = 15x

x = 30/15

x = 2

Get <LOM:

<LOM = 9x+44

<LOM = 9(2)+44

<LOM = 18+44

<LOM = 62°

Get <MON:

<MON = 6x+3

<MON = 6(2)+3

<MON = 12+3

<MON = 15°

6 0

4 years ago

What is the answer ?

snow_tiger [21]

no 1. is a

no 2. is none of the above

8 0

3 years ago