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o-na [289]
2 years ago
10

Please help, thanks!!!

Mathematics
1 answer:
irga5000 [103]2 years ago
6 0

Problem 4

Draw any right triangle you want. Afterward, draw a perpendicular segment that goes from the 90 degree angle to the hypotenuse. Call this new segment the altitude. It might help to have the hypotenuse laying flat or horizontal.

The length of this altitude can be found using the geometric mean formula. Search out "geometric mean for triangles" (or similar) and you should get a diagram that visually summarizes what is going on. I've provided a screenshot of an example using GeoGebra. See below.

==========================================================

Problem 5

The two special types of right triangles are: the 45-45-90 triangle and the 30-60-90 triangle.

If x is the leg length of the first triangle type mentioned, then x\sqrt{2} is the hypotenuse. We can confirm this using the pythagorean theorem. Recall that the legs of any 45-45-90 triangle are the same, meaning we have an isosceles triangle.

For the second type of triangle, the short leg x leads to the hypotenuse 2x and long leg x\sqrt{3}

==========================================================

Problem 6

If you know say the opposite and adjacent sides of a right triangle, then you can use the tangent ratio because tan = opposite/adjacent.

Sine is used for opposite/hypotenuse, while cosine is used for adjacent/hypotenuse. Those are the main 3 trig functions. Technically there are 3 more, but they are just reciprocals of the previous three.

==========================================================

Problem 7

Use inverse trig functions to find the missing angles if you know the sides. The inverse trig functions have an exponent of -1.

For instance, let's say the triangle has an opposite side of 10 and hypotenuse of 15

The reference angle would be...

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(x) = \frac{10}{15}\\\\x = \sin^{-1}\left(\frac{10}{15}\right)\\\\x \approx 41.81^{\circ}\\\\

==========================================================

Problem 8

Imagine looking completely horizontal at the horizon. The angle in which you are currently looking is 0 degrees. If you look upward, say 10 degrees, then the angle of elevation is 10 degrees. Looking down means we have an angle of depression. These two types of angles help us determine various distances and lengths.

Use the alternate interior angle theorem to help show that angles of depression are congruent to angles of elevation. It will depend on context which type of angle is more useful.

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Part A
Anuta_ua [19.1K]

Answer:

Part A: There are 4 zeros of the polynomial function f(x)

Part B: The zeroes of the polynomial function f(x) are -4, -1, 0, 2 ⇒ C

Step-by-step explanation:

<em>The zeroes of a function are </em><em>the x-coordinates of the point of intersection between the graph of the function and the x-axis</em><em> (x-intercepts) which means </em><em>values of x at y = 0</em>

Part A:

In the given graph

∵ The graph of the function f(x) = -x^{4} - 3x³ + 6x² + 8x intersects the x-axis

   at 4 points

→ That means there are 4 values of x have y = 0

∴ The number of zeroes of the function is 4

∴ There are 4 zeros of the polynomial function f(x)

Part B:

∵ The graph intersects the x-axis at points (-4, 0), (-1, 0), (0, 0), (2, 0)

→ That means the values of x at y = 0 are -4, -1, 0, 2

∴ f(x) = 0 at x = -4, -1, 0, 2

∴ The zeroes of the polynomial function f(x) are -4, -1, 0, 2

4 0
3 years ago
Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F
Svet_ta [14]

Answer:

A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36

Step-by-step explanation:

The equations are:

y = x^{2} + 2x + 3

y = 2x + 12

The two graphs intersect when:

x^{2} + 2x + 3 = 2x + 12

x^{2} = 0

x_{1}  = 3\\x_{2}  = -3

To find the area under the curve for the first equation:

A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx

To find the area under the curve for the second equation:

A_{2} = \int\limits^3__-3}{2x + 12} \, dx

To find the total area:

A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx

Simplifying the equation:

A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).  

3 0
3 years ago
What the heck does this mean lol
dangina [55]

Answer:

Step-by-step explanation:

The significant figures of a number that carry meaningful contribution to its measurement resolution

4 0
3 years ago
Read 2 more answers
A box contains 76 coins, only dimes and nickels. The amount in the box is $4.90. How many dimes and how many nickels are in the
dangina [55]

The number of dimes is 75 coins.

The number of nickels is 1 coin.

<u>Step-by-step explanation:</u>

It is given that,

A box contains 76 coins, only dimes and nickels.

  • The 10 cent coin is called a dime.
  • Let, the number of dimes be x.
  • The 5 cent coin is called a nickel.
  • Let, the number of nickels be y.

<u>This is represented by the system of equations :</u>

x + y = 76  -------(1)

10x + 5y = 4.90  ------(2)

Multiply eq (1) by 5 and subtract eq (2) from eq (1),

 5x + 5y = 380

-<u>(10x + 5y = 4.90)</u>

 <u>-5x       = 375.1</u>

x = 375/5

x = 75 coins.

The number of dimes is 75 coins.

Substitute x=75 in eq (1),

y = 76 - 75

y = 1 coin.

The number of nickels is 1 coin.

6 0
3 years ago
Read 2 more answers
Multiplying and Dividing Rational Expressions<br> What is the product?
PolarNik [594]

rational numbers are always negative when u add irrational numbersStep-by-step explanation:

4 0
3 years ago
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