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2 years ago

14

Decide whether a triangle with side lengths of sqr5, 5, and 5.5 is right, acute, or obtuse.

1 answer:

LUCKY_DIMON [66]2 years ago

4 0

It is acute isosceles triangle

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Best to draw out a factor tree.

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Calculating cos-1 ( help is gladly appreciated :) )

Alekssandra [29.7K]

Answer:

$\frac{3\pi}{4}$

(Assuming you want your answer in radians)

If you want the answer in degrees just multiply your answer in radians by $\frac{180^\circ}{\pi}$ giving you:

$\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi}=\frac{3(180)}{4}=135^{\circ}$ .

We can do this since $\pi \text{ rad }=180^\circ$ (half the circumference of the unit circle is equivalent to 180 degree rotation).

Step-by-step explanation:

$\cos^{-1}(x)$ is going to output an angle measurement in $[0,\pi]$ .

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$\cos(x)=-\frac{\sqrt{2}}{2}$ .

This happens in the second quadrant on the given interval.

The solution to the equation is $\frac{3\pi}{4}$ .

So we are saying that $\cos(\frac{3\pi}{4})=\frac{-\sqrt{2}}{2}$ implies $\cos^{-1}(\frac{-\sqrt{2}}{2})=\frac{3\pi}{4}$ since $\frac{3\pi}{4} \in [0,\pi]$ .

Answer is $\frac{3\pi}{4}$ .

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2 years ago

Which inequality defines the problem below ?

Likurg_2 [28]

Answer:

D

Step-by-step explanation:

so total travel by x bikers = 50 multiplied by x = 50x

According to the question

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3 years ago

50 PTS ANSWER ALL <3333333

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QUESTION 33

The length of the legs of the right triangle are given as,

6 centimeters and 8 centimeters.

The length of the hypotenuse can be found using the Pythagoras Theorem.

${h}^{2} = {6}^{2} + {8}^{2}$

${h}^{2} = 36+ 64$

${h}^{2} = 100$

$h = \sqrt{100}$

$h = 10cm$

Answer: C

QUESTION 34

The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.

The length of the other leg can be found using the Pythagoras Theorem,

${l}^{2} + {33}^{2} = {55}^{2}$

${l}^{2} = {55}^{2} - {33}^{2}$

${l}^{2} = 1936$

$l = \sqrt{1936}$

$l = 44cm$

Answer:B

QUESTION 35.

We want to find the distance between,

(2,-1) and (-1,3).

Recall the distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the values to get,

$d=\sqrt{( - 1-2)^2+(3- - 1)^2}$

$d=\sqrt{( - 3)^2+(4)^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d = 5$

Answer: 5 units.

QUESTION 36

We want to find the distance between,

(2,2) and (-3,-3).

We use the distance formula again,

$d=\sqrt{( - 3-2)^2+( - 3- 2)^2}$

$d=\sqrt{( - 5)^2+( - 5)^2}$

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}$

Answer: D

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2 years ago

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