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3 years ago

Given the following right triangle, what is the measure of angle B?

Mathematics

1 answer:

Pepsi [2]3 years ago

5 0

Answer:

A. 55 degrees

Step-by-step explanation:

We are given a right angled triangle ABC having the horizontal length = 10 units and vertical length = 7 units.

Now, the trigonometric form for angles of a right angled triangle gives,

$\tan B=\frac{opposite}{adjacent}$

i.e. $\tan B=\frac{10}{7}$

i.e. $\tan B=1.42857$

i.e. $B=\arctan 1.42857$

i.e. $B=55.03$

Hence, the measure of angle B is 55°

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Help Please!!!!!!!!!!!!!

Brrunno [24]

Answer:

x = 149/2 or x = 74.5

Step-by-step explanation:

GCF = 16

2x - 144 = 5

2x = 149

x = 149/2 or x = 74.5

6 0

3 years ago

Read 2 more answers

Convert the integral below to polar coordinates and evaluate the integral.

Lelechka [254]

Answer:

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

Step-by-step explanation:

We are trying to evaluate this integral.

$\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}} xy \,\,dxdy$

The first thing that we have to do is understand this region in the plane.

$\{ (x,y) \in \mathbb{R} : 0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2} \}$

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

$\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4} \,\,\, 0\leq r \leq 4 \}$

Now remember that when we do the polar transformation we use the following formula

$\int\limits_{a}^{b} \, \int\limits_{c}^{d} f(x,y) \,dxdy = \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2} r* f(rcos(\theta),rsin(\theta)) \,drd\theta$

Then our integral would be

$\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}} xy \, dxdy = \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4} r^3 cos(\theta)\sin(\theta) \,\, drd\theta = 16$

6 0

3 years ago

What is an equation that is perpendicular to the line y = -2?

finlep [7]

Answer:

y=2 is a line parallel to the x axis that passes through the point (0,2)

A line perpendicular to it would any line that is parallel to the y axis, which crosses the x axis at any point.

Step-by-step explanation:

Perpendicular lines are always found by reciprocating the negative value of the slope in question.

The slope in this case, in y = 2, is zero. You have a horizontal line hovering at the value “2” in the y-dimension, parallel to the x dimension. You can tell the slope is zero since there is no coefficient value paired with “x” so you can assume that since value multiplied by 0 is zero, the same instance has been performed in favor of a dearth of “x.”

Alright. So slope zero. The negative of zero is zero, since zero is neutral. It is neither positive nor negative (though some people tend to see it as positive for reasons irrelevant to your question. Will be answered at your request).

If you take the reciprocal of zero, it becomes undefined because what was once a numerator (0 divided by any number except 0, because then it would already be undefined) becomes a denominator (any numerator divided by 0 becomes undefined by default, as explained already). So now you can assume that the new line is undefined. This means that it is vertical! The only line that does not pass as a function. And it makes sense. A slope of 0 is 90° from a slope of no defined value, which will take on an unknown x-value since its location is not specified.

7 0

3 years ago

8 - 10 is the same as

liubo4ka [24]

B. 8+(-10)

Explanation:

8-10 is -2
If you try all the answers, 8+-10=-2

6 0

3 years ago

K+ 6 ≥ 19, if k = 11

WINSTONCH [101]

Answer:

Then it would be the other way around 19 ≥ k + 6

Wait for more responses if needed.