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3 years ago

The legs of a 45-45-90 triangle have a length of 8 units. What is the length of one of its hypotenuse?

1 answer:

8 0

To find the hypotenuse of a 45 45 90 triangle, multiply the leg by sq root (2)

8 * sqrt(2) = 8 * 1.4142135624 
= 11.313708499 
Source:
http://www.1728.org/trig2.htm

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Which graph represents the function g(x) = |x + 4| +2

insens350 [35]

Answer:

none of the two graphs in the pic.

it should be one in the II quadrant

Step-by-step explanation:

the plus four means to move 4 to the left

the plus 2 means to move up 2

That would make the vertex in the II quadrant.

7 0

2 years ago

Read 2 more answers

Two of the exterior angles of a triangle are $158^\circ$ and $99^\circ.$ Find the third exterior angle, in degrees.

Lisa [10]

Answer:

The third exterior angle is $103^o$

Step-by-step explanation:

Given

$\theta = 158^o$

$\alpha = 99^o$

Required

The third exterior angle $(\beta)$

The exterior angles of a triangle add up to 360 degrees.

So:

$\theta + \alpha + \beta = 360^o$

Make $\beta$ the subject

$\beta = 360^o - (\theta + \alpha)$

Substitute known values

$\beta = 360^o - (158^o + 99^o)$

$\beta = 103^o$

4 0

2 years ago

Find the missing side (X) round to the nearest hundredth

nata0808 [166]

Answer:

$\displaystyle 18,28 ≈ x$

Step-by-step explanation:

$\displaystyle \frac{x}{11} = csc\:37 \hookrightarrow 11csc\:37 = x \hookrightarrow 18,278041552... = x \\ \\ 18,28 \approx x$

OR

$\displaystyle \frac{11}{x} = sin\:37 \hookrightarrow xsin\:37 = 11 \hookrightarrow \frac{11}{sin\:37} = x \hookrightarrow 18,278041552... = x \\ \\ 18,28 \approx x$

Information on trigonometric ratios

$\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ$