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3 years ago

6

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!!

1 answer:

gavmur [86]3 years ago

6 0

$\text{Use}\\\\a^2-b^2=(a-b)(a+b)\\\\i=\sqrt{-1}\to i^2=-1\\---------------------\\\\x^2+25=x^2+5^2=x^2-(-1)(5^2)=x^2-(i^2)(5^2)=x^2-(5i)^2\\\\\boxed{x^2+25=(x-5i)(x+5i)}$

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The equation y=-7x-5 and the solution of (-2,9)

Black_prince [1.1K]

Y=-7x-5
9=-7(-2)-5
9=14-5
9=9
Please mark the brainliest, if you think i helped!

8 0

3 years ago

What is 184.24 to 4 significant figures

siniylev [52]

184.2 is the answer to 4 significant figure

4 0

2 years ago

baherus [9]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: cos 330 = $\frac{\sqrt3}{2}$

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

$\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}$

Proof LHS → RHS:

LHS cos 165

Double-Angle: cos (2 · 165) = 2 cos² 165 - 1

⇒ cos 330 = 2 cos² 165 - 1

⇒ 2 cos² 165 = cos 330 + 1

Given: $2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1$

$\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}$

Divide by 2: $\cos^2 165 = \dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}$

Square root: $\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}$

Scratchwork: $\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2$

$\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}$

Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

$\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}$

LHS = RHS $\checkmark$

4 0

3 years ago

Color. Frequency

tankabanditka [31]

Answer:

the answer is blue

6 0

3 years ago

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,

grandymaker [24]

Solution:

There is no saddle point (DNE). However, there is local maximum at (1, 1/2) for the given function.

Explanation:

we have function of two variables f(x,y)= 9-2x+4y-x^2-4y^2

we will find the values by partial derivative with respect to x,y,xy

$f_{x}$ = -2 -2x

$f_{y}$ = 4 -8y

to find the saddle point we should first find the critical points so equate

-2 -2x=0 and 4 -8y=0

we get x= 1 and y =1/2 so, critical points are (1,1/2)

to find local maximum or minimum we have to find $f_{xx}$ , $f_{yy}$ and $f_{xy}$

formula is $f_{xx}$ * $f_{yy}$ - $f^{2_{xy} }$ =0

$f_{xx}$ = -2

$f_{yy}$ = -8

$f^{2_{xy} }$ =0

putting values in formula

(-2)*(-8) -0 =16 > 0, and $f_{xx}$ < 0 and $f_{yy}$ <0

so, here we have local maximum

we have no saddle point for this function by using the same formula we used to find extrema.

4 0

3 years ago

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