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3 years ago

12

Ronnie and Roxanne went to the track after school. Ronnie ran 6 laps in 15 minutes. Roxanne ran 9 laps in 20 minutes. Who ran fa

ster?

2 answers:

stich3 [128]3 years ago

5 0

Roxanne (limit) yeah

kompoz [17]3 years ago

3 0

Answer:

Ronnie ran faster.

Good luck hun!

Step-by-step explanation:

Ronnie;

15/6 = 2.5 laps a minute.

Roxanne;

20/9 = 2.222... laps a minute.

(Ronnie had slightly more laps in a minute than Roxanne did.)

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The vertices of a triangle are p(3,7) Q(7,-6) R(-8,7). Name the vertices of the image reflected across the x axis

pantera1 [17]

Answer:

<em>P'(3,-7) Q'(7,6) R'(-8,-7).</em>

Step-by-step explanation:

<u>Reflection across the x-axis</u>

Given a point P(x,y), its reflection across the x-axis will map to point P'(x,-y), i.e., the y-coordinate gets inverted.

We are given the vertices of a triangle P(3,7) Q(7,-6) R(-8,7). The vertices of the image reflected across the x-axis are:

P'(3,-7) Q'(7,6) R'(-8,-7).

The new triangle has vertices P'Q'R'.

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2 years ago

No fewer than means

eimsori [14]

Answer:

Greater than or equal to, or D.

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2 years ago

Read 2 more answers

s quadrilateral JKLM the result of a dilation of quadrilateral ABCD by a scale factor of 2? Why or why not?

Likurg_2 [28]

If quadrilateral JKLM has given values, as well as quadrilateral ABCD, it can be concluded from the given values if JKLM is a result of a dilation of ABCD by a scale factor of 2. Dilation factor is used to scale up a given figure. If ABCD has measurements of 1, 2, 3, and 4. Then the measurements of JKLM should be 4, 8, 12, and 16.

4 0

3 years ago

Grade 10 math help please. Please show work

masha68 [24]

Answer:

43.94?

Step-by-step explanation:

I think if you use inverse cos you should get the correct number. So, to find cos, you need to know the adjacent side and the hypotenuse. Here, the reference angle would be the person holding the kite. The distance from the person to the barn is the adjacent side, which is 18. The string of the kite, I assume, would be 25.

cos(∅) = $\frac{18}{25}$

cos(∅) = 0.72

cos⁻¹(0.72) = 43.94

I don't know if that's what you're looking for, but... If it's not correct, lmk so I can rewrite it. :)

4 0

3 years ago

A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

irina [24]

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

$x^{2} + y^{2} = z^{2}$

Differentiating the above equation with respect to time yields,

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

dx/dt = 0 since dock height doesn't change

$y\frac{dy}{dt} = z\frac{dz}{dt}$

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

$\frac{dy}{dt} = \frac{13.93}{13}(2)$

$\frac{dy}{dt} = 2.14$ $ft/s$

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

8 0

3 years ago