1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rzqust [24]
3 years ago
11

Ray QM is the angle bisector of angle NQR. If angle PQN = 7x + 15 and angle NQM = 4x, what is the value of x?

Mathematics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

x = 2.7

Step-by-step explanation:

From the question given above, the following data were obtained:

QM = angle bisector of angle PQR

Angle PQN = 7x + 15

Angle NQM = 4x

x =?

Since QM is the angular bisector, it means that it bisect PQR into two equal angle i.e 45° each.

See attachment for diagram.

Thus, we can obtain the value of x as follow:

7x + 15 + 4x + 45= 90° (angle in a right angle triangle)

7x + 4x + 15 + 45= 90

11x + 60 = 90

Collect like terms

11x = 90 – 60

11x = 30

Divide both side by 11

x = 30/11

x = 2.7

Therefore, the value of x is 2.7

You might be interested in
PLEASE HELP!!!!!! DUE TODAY!!!!!! PLEASEEEEEEEE!!!!!!!!
Sergio039 [100]
Once again it's (2a,0) I took this and got it right
4 0
3 years ago
Read 2 more answers
What is -25.83538 times 29 to the power of three divided by 27.954?
soldi70 [24.7K]
I got −15,045,083.2130533
4 0
2 years ago
The height of a hill, h(x), in a painting can be written as a
matrenka [14]

Answer:

First option: 6\ inches

Step-by-step explanation:

<h3> The complete exercise is: "The height of a hill, h(x), in a painting can be written as a function of x, the distance from the left side of the painting. Both h(x) and x are measured in inches h(x) = -\frac{1}{5}(x)(x -13). What is the height of the hill in the painting 3 inches from the left side of the picture?</h3>

You have the following function provided in the exercise:

h(x) = -\frac{1}{5}(x)(x -13)

You know that h(x) represents the height of the hill (in inches) and "x" represents the distance from the left side of the painting (in inches)

Knowing that you can determine that, if the painting 3 inches from the left side of the picture, the value of "x" is the following:

x=3

Therefore, you need to find the value of   h(x) when  x=3 in order to solve this exercise.  

So, the next step is to substitute  x=3 into the function:

h(x) = -\frac{1}{5}(3)(3 -13)

And finally, you must evaluate in order to find h(3).

You get that this is:

h(3) = -\frac{1}{5}(3)(-10)\\\\h(3) = -\frac{1}{5}(-30)\\\\h(3)=6

4 0
3 years ago
Read 2 more answers
BRAINLIESTTTTTTTTTTTT!! :)
skad [1K]
Your answer here would be letter A because for each year it is multiplied by 5 and so year 6 was 31,250 and 31,250 * 5 is 156,250


Have A Good Day :)
3 0
3 years ago
Read 2 more answers
Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
zvonat [6]

Answer:

The maximum volume of a box inscribed in a sphere of radius r is a cube with volume \frac{8r^3}{3\sqrt{3}}.

Step-by-step explanation:

This is an optimization problem; that means that given the constraints on the problem, the answer must be found without assuming any shape of the box. That feat is made through the power of derivatives, in which all possible shapes are analyzed in its equation and the biggest -or smallest, given the case- answer is obtained. Now, 'common sense' tells us that the shape that can contain more volume is a symmetrical one, that is, a cube. In this case common sense is correct, and the assumption can save lots of calculations, however, mathematics has also shown us that sometimes 'common sense' fails us and the answer can be quite unintuitive. Therefore, it is best not to assume any shape, and that's how it will be solved here.

The first step of solving a mathematics problem (after understanding the problem, of course) is to write down the known information and variables, and make a picture if possible.

The equation of a sphere of radius r is x^2 + y^2 + z^2=r^2. Where x, y and z are the distances from the center of the sphere to any of its points in the border. Notice that this is the three-dimensional version of Pythagoras' theorem, and it means that a sphere is the collection of coordinates in which the equation holds for a given radius, and that you can treat this spherical problem in cartesian coordinates.

A box that touches its corners with the sphere with arbitrary side lenghts is drawn, and the distances from the center of the sphere -which is also the center of the box- to each cartesian axis are named x, y and z; then, the complete sides of the box are measured  2x,  2y and 2z. The volume V of any rectangular box is given by the product of its sides, that is, V=2x\cdot 2y\cdot 2z=8xyz.

Those are the two equations that bound the problem. The idea is to optimize V in terms of r, therefore the radius of the sphere must be introduced into the equation of the volumen of the box so that both variables are correlated. From the equation of the sphere one of the variables is isolated: z^2=r^2-x^2 - y^2\quad \Rightarrow z= \sqrt{r^2-x^2 - y^2}, so it can be replaced into the other: V=8xy\sqrt{r^2-x^2 - y^2}.

But there are still two coordinate variables that are not fixed and cannot be replaced or assumed. This is the point in which optimization kicks in through derivatives. In this case, we have a cube in which every cartesian coordinate is independent from each other, so a partial derivative is applied to each coordinate independently, and then the answer from both coordiantes is merged into a single equation and it will hopefully solve the problem.

The x coordinate is treated first: \frac{\partial V}{\partial x} =\frac{\partial 8xy\sqrt{r^2-x^2 - y^2}}{\partial x}, in a partial derivative the other variable(s) is(are) treated as constant(s), therefore the product rule is applied: \frac{\partial V}{\partial x} = 8y\sqrt{r^2-x^2 - y^2}  + 8xy \frac{(r^2-x^2 - y^2)^{-1/2}}{2} (-2x) (careful with the chain rule) and now the expression is reorganized so that a common denominator is found \frac{\partial V)}{\partial x} = \frac{8y(r^2-x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}  - \frac{8x^2y }{\sqrt{r^2-x^2 - y^2}} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}.

Since it cannot be simplified any further it is left like that and it is proceed to optimize the other variable, the coordinate y. The process is symmetrical due to the equivalence of both terms in the volume equation. Thus, \frac{\partial V}{\partial y} = \frac{8x(r^2-x^2 - 2y^2)}{\sqrt{r^2-x^2 - y^2}}.

The final step is to set both partial derivatives equal to zero, and that represents the value for x and y which sets the volume V to its maximum possible value.

\frac{\partial V}{\partial x} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} =0 \quad\Rightarrow r^2-2x^2 - y^2=0 so that the non-trivial answer is selected, then r^2=2x^2+ y^2. Similarly, from the other variable it is obtained that r^2=x^2+2 y^2. The last equation is multiplied by two and then it is substracted from the first, r^2=3 y^2\therefore y=\frac{r}{\sqrt{3}}. Similarly, x=\frac{r}{\sqrt{3}}.

Steps must be retraced to the volume equation V=8xy\sqrt{r^2-x^2 - y^2}=8\frac{r}{\sqrt{3}}\frac{r}{\sqrt{3}}\sqrt{r^2-\left(\frac{r}{\sqrt{3}}\right)^2 - \left(\frac{r}{\sqrt{3}}\right)^2}=8\frac{r^2}{3}\sqrt{r^2-\frac{r^2}{3} - \frac{r^2}{3}} =8\frac{r^2}{3}\sqrt{\frac{r^2}{3}}=8\frac{r^3}{3\sqrt{3}}.

6 0
3 years ago
Other questions:
  • The sum of two numbers is 15 and their product is 54.What are the two numbers?
    6·1 answer
  • I NEED HELP WITH THIS!! PLEASE HELP ME I NEED IT ASAP<br> 12 POINTS
    11·1 answer
  • Suppose you're given the formula P = 9m – 3n. If you know that n is 1⁄3 of m, how could you rewrite this formula?
    14·2 answers
  • What is the solution of the equation of the compound inequality 4x &lt;-8 or 9x&gt; 18
    9·1 answer
  • Marie is a party planner who is helping a couple plan their wedding reception. Marie suggested 2 different reception locations a
    14·1 answer
  • Select ALL the correct answers. Select all the expressions that are equivalent to the polynomial below. 4(x − 4) − 2(3x2 + 3x −
    15·2 answers
  • PACK is a rectangle with a perimeter of 40 units. Find the measure of angle O to the nearest degree.
    12·1 answer
  • M is the midpoint of AB has coordinates(5, −4), if A has coordinates(6,1). What are thecoordinates of B. (1.5pts)?
    5·1 answer
  • The hypotenuse of a right triangle is 5 meters longer than one of its legs. The other leg is 6 meters. Find the length of the ot
    12·1 answer
  • Find x and y for this triangle
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!