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andrew-mc [135]

3 years ago

11

Alex is setting up an in-line skating course 21 feet long to practice weaving around cones. He wants a cone every 3 1/2 feet. Ho

w many cones does he need?

1 answer:

Sauron [17]3 years ago

3 0

Answer:

I think it is 6 cones

Step-by-step explanation:

21 ÷ 3.5 = 6

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Can some one plz help fast F=-m+B/q^3

Pani-rosa [81]

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5 0

3 years ago

A small pebble has a mass of<br> about<br> 20 L<br> b.<br> 20 ml<br> 20 g<br> 20 kg<br> d.

zhuklara [117]

Answer:

20g

since mass should be in kg or gram if small then g

7 0

3 years ago

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-3,1) and (2,-1) is $y=-\frac{2}{5}x-\frac{1}{5}$

8 0

3 years ago

Please help 10 points and I will be ur friend I need help with Geometry bad

sineoko [7]

Ok, let's use N and N'.
N is (-1,-1). N' is (4,6).
From point N, the quadrilateral moves to the right 5 and up 7.
So, N+(5,7)=N'.

3 0

3 years ago

Solve the following system of equations. What is one x-value of a solution?

gulaghasi [49]

Answer:

There is some mistake in the question, because the solutions are x = -1.445 and x = -34.555

Step-by-step explanation:

Given the functions:

f(x) = x² + 4x + 10

g(x) = -32x - 40

we want to find the points at which f(x) = g(x).

x² + 4x + 10 = -32x - 40

x² + 4x + 10 + 32x + 40 = 0

x² + 36x + 50 = 0

Using quadratic formula:

$x = \frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)}$

$x = \frac{-36 \pm \sqrt{36^2-4(1)(50)}}{2(1)}$

$x = \frac{-36 \pm 33.11}{2}$

$x_1 = \frac{-36 + 33.11}{2}$

$x_1 = -1.445$

$x_2 = \frac{-36 - 33.11}{2}$

$x_2 = -34.555$

5 0

3 years ago