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3 years ago

I need to know the distance and if your could please tell me the formula for future problems thank you!! ASAP

Mathematics

1 answer:

Ksivusya [100]3 years ago

8 0

Distances in 2- and 3-dimensions (and even higher dimensions) can be found using the Pythagorean theorem. The straight-line distance can be considered to be the hypotenuse of a right triangle whose sides are the horizontal and vertical differences between the coordinates.

Here, you have A = (0, 0) and B = (3, 6). The horizontal distance between the points is ...

... 3 - 0 = 3 . . . . the difference of x-coordinates

The vertical distance between the points is ...

... 6 - 0 = 6 . . . . the difference of y-coordinates

Then the straight-line distance (d) between the points is found from the Pythagorean theorem, which tells you ...

... d² = 3² + 6²

... d = √(9 + 36) = √45 ≈ 6.7 . . . units

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Solve the equation for x and enter your answer in the box below x + 14 =27

Helen [10]

Answer: x = 13

Step-by-step explanation: When solving an equation like this, we are trying to get our variable which is our letter by itself.

So we first want to ask ourselves what is the 14 doing to x. Well, we can see that it's being added to x so to get x by itself, we will do the opposite of addition which is subtraction. So we subtract 14 from both sides of the equation.

The +14 -14 cancels out so we're left with x on the left.

On the right, we must subtract 14 from 27 to get 13.

So we have x = 13 which is the solution to this equation.

3 0

3 years ago

Read 2 more answers

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2