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nikitadnepr [17]
3 years ago
7

What is the solution(s) to the equation (2x-6)(3x-4)=0?

Mathematics
1 answer:
Strike441 [17]3 years ago
8 0
Why my answer needs to be 20 characters long... what if it is only 19 characters??

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In the standard (x,y) coordinate plane, what is the slope of the line 4x+7y=9?​
o-na [289]

Answer: -4/7

Step-by-step explanation:

To find the slope, let's change the equation to slope-intercept form.

4x+7y=9              [subtract both sides by 4x]

7y=-4x+9           [divide both sides by 7]

y=-\frac{4}{7}x+\frac{9}{7}

Now, we know the slope is -4/7.

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3 years ago
What is (7÷2+2×2×)×10
bulgar [2K]
The answer is 75

I hope this helps. 

Have a great day. :)
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2 years ago
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What is area of this
nata0808 [166]

Answer:

A = 20.5 sq units

Step-by-step explanation:

you have the height of 5 but need to figure out the base

we can use the Pythagorean Theorem two times to find it

1) a² + 5² = 7²

a² = 49-25

a = \sqrt{24} = \sqrt{4}·\sqrt{6} = 2\sqrt{6}

2) b² + 5² = 6²

b² = 36-25

b = \sqrt{11}

Base = 2\sqrt{6} + \sqrt{11}

A = 1/2 x 5· (2\sqrt{6} + \sqrt{11})

A = 2.5 · (2\sqrt{6} + \sqrt{11})

A = 2.5(8.2)

A = 20.5

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The correct answer is A
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3 years ago
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Find all values of the angle θ (in radians, with 0 ≤ θ < 2π) for which the matrix a = cos θ −sin θ sin θ cos θ has real eigen
harina [27]

The matrix

A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}

has eigenvalues \lambda such that

\det(A-\lambda I)=\begin{vmatrix}\cos\theta-\lambda&-\sin\theta\\\sin\theta&\cos\theta-\lambda\end{vmatrix}=0

(\cos\theta-\lambda)^2+\sin^2\theta=0

(\cos\theta-\lambda)^2=-\sin^2\theta

\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}

\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}

\sin^2\theta\ge0 for all values of \theta, so we need to have \sin\theta=0 in order for \lambda to be real-valued. This happens for

\sin\theta=0\implies\theta=n\pi

where n is any integer, and over the given interval we have \theta=0 and \theta=\pi.

4 0
3 years ago
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