Identify the cross section that results from slicing the three-dimensional figure with a plane that is parallel to its base

Solve this quadratic equation by completing the square. x2 + 10x = 26

Misha Larkins [42]

  x² + 10x = 26
  x² + 10x + 25 = 26 + 25
  x² + 5x + 5x + 25 = 51
x(x) + x(5) + 5(x) + 5(5) = 51
x(x + 5) + 5(x + 5) = 51
(x + 5)(x + 5) = 51
(x + 5)² = 51
x + 5 = ±√(51)
- 5 - 5
x = -5 ± √(51)
x = -5 + √(51) or x = -5 - √(51)

5 0

3 years ago

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How to solve the problem

morpeh [17]

Might be better if you typed the question

8 0

4 years ago

Please help<br><br> 3(-21 + k) = 18

Novay_Z [31]

Answer:

K=27

Step-by-step explanation:

3(-21+k)=18

-63+3k=18

3k=18+63

3k=81

K=81/3

K=27

7 0

3 years ago

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Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =

Black_prince [1.1K]

Answer:

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line $x + 2y= 0, x\geq 0$

If we choose $x = 2$ (we can do so because of the condition $x \geq 0$ , which means that any positive value is suitable for $x$ ), then we have

$2 +2y = 0\implies 2 = -2y \implies y = -1$

Therefore, the terminal side of the angle $\theta$ is passing through the origin and the point $(2,-1)$ and now we can draw it.

The angle $\theta$ is presented below.

The distance of the point $(2,-1)$ from the origin equals

$r = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Now, we can determine the values of the six trigonometric function, by using their definitions.

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

6 0

3 years ago

SOMEONE, PLEASE HELP, I NEED THE ANSWER NOW

faltersainse [42]

3. -155.15
4. -140.95
5. -1.40
6. 5.80
I’m not 100% sure but I think it’s right

8 0

3 years ago