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PtichkaEL [24]
2 years ago
12

How do i factor equations?

Mathematics
1 answer:
Pavel [41]2 years ago
7 0

Answer:

If your quadratic equation it is in the form of x^2+bx+c=0 (in other words, if the coefficient of the x^2 term=1), it's possible (but not guaranteed) that a relatively simple shortcut can be used to factor an equation. Find two numbers that both multiply to make C and add to make B.

Step-by-step explanation:

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Look at the picture......
Mashcka [7]

Answer:

the coordinates would be (3,6)

5 0
3 years ago
Identify a pattern by making a table of the inputs and outputs.
-BARSIC- [3]

it wants you to make a table of the inputs and out puts, and what is the pattern that is happening. like is it a constant increase/decrease and things like that. i'll help you on the table...

x | y

1 | 5

2 | 4

3 | 3

4 | 2

(i am not sure what that 40 is at the end, but You get the point)

now all you have to do is identify the sequence or pattern you see here


3 0
3 years ago
2sin^(-2490)+tan 1410
givi [52]
2sin²(-2490)+tan 1410=2sin²(-330-6*360)+tan (330+3*360)=
=2sin²(-330)+tan (330)=2 sin²(30-1*360)+tan (-30)=
=2sin²(30)-tan (30)=2(1/2)²-√3/3=1/2-√3/3)=(3-2√3)/6

Answer: 2 sin²(-2490)+tan (1410)=(3-2√3)/6    (≈-0.07735...)
7 0
3 years ago
(2a^3 b^4) ^4= <br> What is the answer to this problem ?
Jet001 [13]

Answer: 16a^12 b^16

Step-by-step explanation:

Everything in the parenthesis to the 4th power.

2a^4= 16a

3*4=12

b^4*4=b^16

3 0
2 years ago
If
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
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