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Debora [2.8K]
2 years ago
15

I need help! This is Geometry lesson 8-3 ! Please help me asap!!!!

Mathematics
1 answer:
gulaghasi [49]2 years ago
8 0
If you learned about the 45-45-90 triangle (which is isosceles), then the faster way is to know that the hypotenuse (side opposite of right angle) is √2 times either one of the sides.
3√2 = (√2)x
x = 3

But if you didn't learn the 45-45-90 triangle yet, that's ok.
Recall the trigonometric ratios for right triangles: sine (sin), cosine (cos), tangent (tan).

If your angle is x, then
sin(x) = opposite side / hypotenuse
cos(x) = adjacent side / hypotenuse
tan(x) = opposite side / adjacent side

Remember the hypotenuse is the side opposite and across from the right angle (3√2 in this case).
An acronym to remember this is SohCahToa.

In this problem, the angle given is 45°, and you need to find the length of the adjacent side x. The hypotenuse is also given as 3√2.

Because we have the adjacent side and the hypotenuse, we use cosine to relate those two sides
cos(45°) = x / (3√2)
x = (3√2)cos45°
If you plug this into your calculator (in degree mode), then
x = 3

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How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?
RSB [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Let us consider

y = f(x)

To find the area under the curve y = f(x) between x = a and x = b, all we need is to integrate y = f(x) between the limits of a and b.

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

A=\int _a^b|f\left(x\right)|dx

    = \int _{-2}^2\left|x^2-4\right|dx

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solving

\int _{-2}^2x^2dx

\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

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computing the boundaries

     =\frac{16}{3}

Thus,

\int _{-2}^2x^2dx=\frac{16}{3}

similarly solving

\int _{-2}^24dx

\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax

     =\left[4x\right]^2_{-2}

computing the boundaries

      =16

Thus,

\int _{-2}^24dx=16

Therefore, the expression becomes

A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx

  =\frac{16}{3}-16

  =-\frac{32}{3}

  =-10.67 square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.

   

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Answer:

Step-by-step explanation:

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-x + 2y = -15

-------------------add

8y = - 6 <=== u see that the x terms cancel each other out

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Please help!
siniylev [52]

Answer:

The answer is 72

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