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inessss [21]
3 years ago
7

Which graph represents the solution to the system of inequalities? x2 - 4y2 < 16 y< 2x

Mathematics
1 answer:
wel3 years ago
8 0
The graph is 6-3 over 8 yw
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Select each perfect square
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hi

Step-by-step explanation:

1 and 2 no are perfect square other r not

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Describe relationship I WILL MARK BRAINLIEST IF YOU GET IT RIGHT
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cleetus

Step-by-step explanation:

5 0
3 years ago
What does this mean??4/9 ÷ 17/ 18
vichka [17]

Answer:

4/9 ÷ 17/18 =

Step-by-step explanation:

4/9 ÷ 17/18

4/9 · 18/17      (Take the reciprocals of 17/18)

72/153

Simplest form : 8/17

Hope this helps!

-Abha

8 0
3 years ago
Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
polet [3.4K]
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}

The the matrix product becomes:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
8 0
3 years ago
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An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle
omeli [17]

Answer:

\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}

Step-by-step explanation:

The length of the base is the distance between the points 4+2i and 10+4i, so

\text{Base}=|10+4i-(4+2i)|=|10+4i-4-2i|=|6+2i|=\sqrt{6^2+2^2}=\\ \\=\sqrt{36+4}=\sqrt{40}

The middle point of the base is placed at point

\dfrac{4+2i+10+4i}{2}=\dfrac{6i+14}{2}=7+3i

The length of the height is the distance between the points 5+9i and 7+3i

\text{Height}=|5+9i-(7+3i)|=|5+9i-7-3i|=|-2+6i|=\sqrt{(-2)^2+6^2}=\\ \\=\sqrt{4+36}=\sqrt{40}

So, the area of the triangle is

A=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}=\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}

4 0
3 years ago
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