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Kobotan [32]
2 years ago
12

Hey guys back again ,If you have any doubt in maths you can ask ​

Mathematics
1 answer:
Juli2301 [7.4K]2 years ago
3 0
Answer
What is acute angle
Step-by-step explanation:

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Solve 6 over x minus 3 equals 3 over x for x and determine if the solution is extraneous or not
Natali5045456 [20]

Answer:

<em>x = 1</em>

This is the equation that I interpreted:

6/(x) - 3 = 3/(x)

I have never worked with determining extraneous equations before, so I cannot answer the second part.

8 0
3 years ago
Write the equation of the circle graphed below.
Darina [25.2K]

Answer:

Your equation is (x - 1)^2 + (y +1)^2 = (\frac{1}{2}  )^2

Step-by-step explanation:

Well, the center origin of the circle is given (h,k) =  (1,-1).

We have to find our radius as they gave us a point. from origin to the edge of the circle.

Using the formula: (x - h)^2 + (y - k)^2 = r^2

Plug in our (h,k) = (1,-1) and (x,y) =  (0.5,-1) to solve for radius.

(x - h)^2 + (y - k)^2 = r^2

(0.5 - (1))^2 + (-1 - (-1))^2 = r^2

(-0.5)^2 + (0)^2 = r^2

1/4= r^2

r^2 = 1/4

r = 1/2

6 0
2 years ago
5/8 didved by 1/4 plz help me with this
Vedmedyk [2.9K]

Answer:

The answer is 2.5 or 5/2 as a fraction.

Step-by-step explanation:

P.S Can I have brainliest?

5 0
3 years ago
Read 2 more answers
What is the Slope of the line
vampirchik [111]

Answer:

m=\frac{-3}{4}

Step-by-step explanation:

Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Simply plug in 2 coordinates into the slope formula to find slope <em>m</em>:

m=\frac{-4-(-1)}{5-1}

m=\frac{-4+1}{4}

m=\frac{-3}{4}

5 0
3 years ago
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
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