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ankoles [38]
4 years ago
10

Subtract (-2x^2+9x-3)-(7x^2-4x+2)

Mathematics
2 answers:
V125BC [204]4 years ago
8 0

Answer: =-9x^2+13x-5

Step-by-step explanation:

First, you need to remember the multiplication of signs:

(+)(+)=+\\(-)(-)=+\\(-)(+)=-

Then, to subtract the polynomials given, the first step is to distribute the negative sign:

(-2x^2+9x-3)-(7x^2-4x+2)=-2x^2+9x-3-7x^2+4x-2

And finally, you need to add the like terms.

With this procedure,  you get the following result:

=-9x^2+13x-5

Fynjy0 [20]4 years ago
6 0

For this case we must subtract the following expression:

(-2x ^ 2 + 9x-3) - (7x ^ 2-4x + 2) =

We must bear in mind that:

- * + = -\\- * - = +

We rewrite the expression:

-2x ^ 2 + 9x-3-7x ^ 2 + 4x-2 =

We add similar terms taking into account that equal signs are added and the same sign is placed, while different signs are subtracted and the sign of major is placed:

-2x ^ 2-7x ^ 2 + 9x + 4x-3-2 =\\-9x ^ 2 + 13x-5

Answer:

-9x ^ 2 + 13x-5

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Please prove this........​
Crazy boy [7]

Answer:  see proof below

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

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

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4 years ago
a restaurant can seat 100 people . it has booths that seat 4 people and tables that seat 6 people.so far , 5 of the booths are f
Natasha2012 [34]
Is this the whole question? If so, this how I found my answer(I do not know if this is the answer 100% and I won't and can't take credit for the answer being incorrect. It is your choice to go with my answer, thank you.): 

A restaurant can seat 100 people. Okay, the capacity is 100. Got it. It has booths that seat 4 people. Okay! So I am going to represent B"Booths" as "B". A variable. B = 4. Okay, let's continue. And, it has tables that seat 6 people. I will represent "Tables" as T. So, T = 6!

Alllrighty then, let's get to the nit and grit..

B = 4 T = 6

In the last sentence, it says, 5 of the booths are full, what expression matches the situation? Well, it's ONLY asking for the expression, so that's all we'll give.

We don't need the "table" variable because it's asking about the booths so, bye bye tables. B = 4, and 5 of them are full. We would have to do B(5) or B * 5. B equals 4 so 4 * 5 which equals 20. The expressions we could use are:

B(5) = 20
 or
4 * 5 = 20
or
20 = 4 * 5 = B(5)

I hope this answer helps you out! Bye! :)
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Answer:

don't know lol my boy hope this helps

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