All categories

3 years ago

What is the distance between nicholas's school and the post office?

1 answer:

4 0

Part A:

Given that Nicholas's school is located at point (3, 2) and that the post office is located at point (-2, -2), then the distance between Nicholas's school and the post office is given by:

$d= \sqrt{(-2-3)^2+(-2-2)^2} \\ \\ = \sqrt{(-5)^2+(-4)^2} = \sqrt{25+16} \\ \\ \sqrt{41} =\bold{6.4} \ units$

Part B:

If Nicholas is located at point (– 3 , 2), the distance to get to the grocery store located at point (1,-1) is given by:

$d= \sqrt{(1-(-3))^2+(-1-2)^2} \\ \\ = \sqrt{(1+3)^2+(-3)^2} = \sqrt{(4)^2+9} \\ \\ = \sqrt{16+9} = \sqrt{25} =\bold{5}$

You might be interested in

Which functions are symmetric with respect to the origin?

Damm [24]

Answer:

The correct option is;

y = arcsinx and y = arctanx

Step-by-step explanation:

The given options are;

1) y = arcsinx and y = arccosx

Here, we have at the origin, where x = 0, arccosx ≈ 1.57 while arcsinx = 0

Therefore arccosx does not intersect arcsinx at the origin for it to be symmetrical to arcsinx or the origin

2) y = arccosxy and y = arctanx

Here arctanx = 0 when x = 0 and arcos x = 1.57 when x = 0 therefore, they are not symmetrical

3) y = arctanx and y = arccotx

Similarly, At x = 0, arccotx = 1.57 therefore, they are not symmetrical

4) y = arcsinx and y = arctanx

Both functions arcsinx and arctanx pass through the origin and their shapes are similar but inverted as they go from negative to positive.

3 0

3 years ago

You flip a coin 10 times and find the experimental probability of flipping tails to be 0.7. Does this seem reasonable? Explain.

Vedmedyk [2.9K]

No. I would say 50/50 chance. This answer probably isnt helpful :/

8 0

3 years ago

3x + 1 = 5X-13 if anyone knows what to do can they please help

Elza [17]

Answer:

x = 7

Step-by-step explanation:

First, you look at the equation. Identify the locations of variable terms (both sides of the equal sign) and the constant terms (both sides of the equal sign).

If there are any parentheses, it is a good idea to use the distributive property to eliminate them. Here, there are none.

I like to start by subtracting <em>the variable term with the smallest coefficient</em>. Here, that is 3x, so we add -3x to both sides of the equation.

3x -3x +1 = 5x -3x -13

1 = 2x -13 . . . . . . . . . . . . combine like terms

Now, we have the only variable term on one side of the equal sign. We want it by itself, so we need to make the -13 go away. We do that by adding its opposite to both sides of the equation:

1 +13 = 2x -13 +13

14 = 2x . . . . . . . . . . . . combine like terms

Finally, we want the coefficient of 2 in the x-term to disappear. We make that happen by multiplying both sides of the equation by 1/2, the reciprocal of that coefficient.

(1/2)(14) = (1/2)(2x)

7 = x . . . . the solution

It is generally a good idea to <em>check your work</em> by seeing if your solution value satisfies the equation:

3(7) +1 = 5(7) -13 . . . . put 7 where x is in the original equation

21 +1 = 35 -13

22 = 22 . . . . . . x = 7 is the solution

_____

<em>Additional comment</em>

By subtracting 3x from 5x, the result is 2x with a positive coefficient. We could solve the equation just as easily by subtracting 5x from 3x. That result would be ...

-2x +1 = -13

Subtracting 1 would give

-2x = -14

and you would multiply by -1/2 to get x=7. I personally like to avoid having this many minus signs show up in the problem. That is why I choose to subtract the x-term with the smallest coefficient.

5 0

3 years ago

Which line is parallel to the line<br> shown below?

Travka [436]

<h3>Answer: Choice D</h3>

4x - 3y = 15

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

Explanation:

The two points (-1,-1) and (2,3) are marked on the line

Let's find the slope of the line through those two points.

$(x_1,y_1) = (-1,-1) \text{ and } (x_2,y_2) = (2,3)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{3 - (-1)}{2 - (-1)}\\\\m = \frac{3 + 1}{2 + 1}\\\\m = \frac{4}{3}\\\\$

The slope is 4/3 meaning we go up 4 and to the right 3.

-------------

Parallel lines have equal slopes, but different y intercepts. We'll need to see which of the four answer choices have a slope of 4/3.

Solve the equation in choice A for y. The goal is to get it into y = mx+b form so we can determine the slope m.

$3x + 4y = -4\\\\4y = -3x-4\\\\y = -\frac{3}{4}x-\frac{4}{4}\\\\y = -\frac{3}{4}x-1$

Equation A has a slope of -3/4 and not 4/3 like we want.

Therefore, this answer choice is crossed off the list.

Follow similar steps for choices B through D. I'll show the slopes of each so you can check your work.

slope of equation B is 3/4
slope of equation C is -4/3
slope of equation D is 4/3

We have a match with equation D. Therefore, the equation 4x-3y = 15 is parallel to the given line shown in the graph.

You can use graphing tools like Desmos or GeoGebra to confirm the answer.

4 0

8 months ago

Verify the trigonometric identities

snow_lady [41]

1)

here, we do the left-hand-side

$\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2
\\\\\\\
[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]
\\\\\\
2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2$

2)

here we also do the left-hand-side

$\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)
\\\\\\
\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}
\\\\\\
\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)$

3)

here, we do the right-hand-side

$\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}
\\\\\\
\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}
\\\\\\
\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}$

8 0

3 years ago