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3 years ago

13

Help me out please..........

1 answer:

evablogger [386]3 years ago

7 0

I’m not sure for the other ones but D is def. likely

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Can someone help me with this question? I seem to be getting a negative answer every time :(

NeX [460]

Substitute 4 for y. Then -3x^2 + 16 = 52.

Solve for x. Subtract 16 from both sides, obtaining -3x^2 = 36.

Divide both sides by -3, obtaining x^2 = -12. This last result makes no sense, as no square of a real number could be negative. Probably this is where you're ":getting a negative answer."

If imaginary answers were allowed, then x = i*√12 = i*2√3 or x = -i*2√3.

x =

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4 years ago

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Find the angle θ between u = 〈6, –5〉 and v = 〈11, 8〉.

nexus9112 [7]

Step-by-step explanation:

this is the answer in the picture

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3 years ago

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-8x2(x+8)=8(2+6) how do i solve this and 4n+3/2=2n-5/1?

Marat540 [252]

1) $-8x^2\left(x+8\right)=8\left(2+6\right)$

$-8x^3-64x^2=64$

$-8x^3-64x^2-64=64-64$

$-8x^3-64x^2-64=0$

$-8x^2+0.97035\dots x-7.88051\dots \approx \:0$

$x\approx \:-8.12129\dots$

2) $4n+\frac{3}{2}=2n-\frac{5}{1}$

$4n+\frac{3}{2}=2n-5$

$4n+\frac{3}{2}-\frac{3}{2}=2n-5-\frac{3}{2}$

$4n=2n-\frac{13}{2}$

$4n-2n=2n-\frac{13}{2}-2n$

$2n=-\frac{13}{2}$

$\frac{2n}{2}=\frac{-\frac{13}{2}}{2}$

$n=-\frac{13}{4}$

6 0

3 years ago

A(-1, 1)b(1, -2)c(0, -4) rotated 270 degrees by the origin

Ilia_Sergeevich [38]

Use the formula (y,-x)
a: (1,1)
b’(-2,-1)
c’(-4,0)

8 0

3 years ago

Please please please need help on 3 and 4 I’m really lost on them

Alex73 [517]

Answer:

1) Option c) is correct ie., 5 real and o non-real

2) Option b) is correct ie., (4, $\frac{1}{2}$ , $\frac{1}{2}$ , 2,2)

Step-by-step explanation:

Given polynomial function is $f(x)=x^5-3x^3-2$

To find zeros equate f(x) to zero ie., $f(x)=0$

$x^5-3x^3-2=0$

By synthetic division

| 1 0 -3 0 -2

-1 | 0 -1 1 2 2

|_________________

1 -1 -2 2 0

Therefore x=-1 is a zero

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

| 1 -1 -2 2

1 | 0 1 0 2

|___________________

1 0 -2 0

x=1 is the zero

$x^2 -2 =0$

$x=\pm\sqrt{2}$

$x=\sqrt{2}$ and

$x=\sqrt{-2}$

Option c) is correct ie., 5 real and o non-real

2) Given polynomial function is $f(x)=(x-4)(2x-1)^2(x-2)^2$

To find zeros equate f(x) to zero ie., $f(x)=0$

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

$(x-4)=0$ (or) $(2x-1)^2=0$ or $(x-2)^2=0$

Therefore x=4, $x=\frac{1}{2}$ of multiplicity of 2 and x=2 multiplicity of 2

Option b) is correct ie., (4, $\frac{1}{2}$ , $\frac{1}{2}$ , 2,2)

5 0

4 years ago