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Lostsunrise [7]
2 years ago
10

Does anyone know the answer?

Mathematics
1 answer:
Serhud [2]2 years ago
8 0

Answer:

6a {}^{2}  - 2c + 2a {}^{2}  - 2c {}^{2}  \\ 8a {}^{2}  - 2c {}^{2}  - 2c

Good luck!

Intelligent Muslim,

From Uzbekistan.

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A slice is made perpendicular to the base of a right rectangular prism as shown.
Eduardwww [97]

Answer:

The correct option is 1. The area of cross section area is 48 mm².

Step-by-step explanation:

From the find it is noticed that the cross section is a rectangle with length 4 mm and width is 12 mm.

The area of a rectangle is the product of its dimensions.

A=l\times w

Where, l is length of the rectangle and w is width of the rectangle.

The area of cross section is

A=4\times 12

A=48

Therefore the area of cross section area is 48 mm². Option 1 is correct.

8 0
3 years ago
Which of the following expressions has a value between 5 and 6.
Airida [17]
So listen to this logic
lets say you have 3 numbers, m,a,x where they are all positivie
m>a>x
therefor
m^2>a>2>x^2
if m and a and x are square roots, therefor we can reverse the square root
confusing sorry

square 5 and 6
5^2=25
6^2=36

the number betwe 25 and 36

that would be 32

answer is √32

goes like this
5^2<32<6^2
sqrt both sides
5<√32<6

answer is √32
8 0
3 years ago
Read 2 more answers
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

7 0
3 years ago
ANSWER NOW FOR A LOT OF POINTS!
sleet_krkn [62]
Answer: x>2
Explanation:
4x- 2>6
4x>8
x> 2
HOPE IT HELPS!!!:):)
3 0
3 years ago
Graph the system of equations. somebody please help
Brums [2.3K]
More or less is this

8 0
2 years ago
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