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antiseptic1488 [7]
2 years ago
15

All acute triangles are similar. a. True b. False

Mathematics
2 answers:
nirvana33 [79]2 years ago
8 0
Kind of true and kind of false : Sometimes they are, but there is nothing that force two acture triangles to be similar, even if they have the same measurement of the acute angle.

For that you would need to have 3 angles the same.

For example, if the angles were 40-60-80, you have an acute triangle. But if you had a 45-55-80 triangle, that is also acute, but not similar even though it shares an angle of 80 degrees.
Ksju [112]2 years ago
8 0

Answer:

The answer is False.

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A negative number divided by a positive number is:
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Step-by-step explanation:

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7 0
2 years ago
Which statement describes the inverse of m(x) = x^2 – 17x?
DochEvi [55]

Given:

The function is

m(x)=x^2-17x

To find:

The inverse of the given function.

Solution:

We have,

m(x)=x^2-17x

Substitute m(x)=y.

y=x^2-17x

Interchange x and y.

x=y^2-17y

Add square of half of coefficient of y , i.e., \left(\dfrac{-17}{2}\right)^2 on both sides,

x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2        [\because (a-b)^2=a^2-2ab+b^2]

Taking square root on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}

Add \dfrac{17}{2} on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y

Substitute y=m^{-1}(x).

m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}

We know that, negative term inside the root is not real number. So,

x+\left(\dfrac{17}{2}\right)^2\geq 0

x\geq -\left(\dfrac{17}{2}\right)^2

Therefore, the restricted domain is x\geq -\left(\dfrac{17}{2}\right)^2 and the inverse function is m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}.

Hence, option D is correct.

Note: In all the options square of \dfrac{17}{2} is missing in restricted domain.

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The measure of an angle is 6° less than the measure of its complement.
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