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Y_Kistochka [10]

4 years ago

10

A 4f by 5ft dock is anchored in the middle of a lake. The bow of a boat is tied to a corner of the dock with a 5ft rope as shown

in the picture. Find the area of the region in which the bow of the boat can travel. Round your answer to the nearest foot

1 answer:

makvit [3.9K]4 years ago

4 0

Is there a photo or something

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What is the solution?

mrs_skeptik [129]

Answer:

(2,3)

Step-by-step explanation:

it's the point of intersection

hope this helps, pls mark brainliest :D

7 0

3 years ago

Read 2 more answers

Graph the line with a slope of 2/3 that contains the point (−3,−5)

Anna007 [38]

Answer:

Step-by-step explanation:

So for the given information, it would be easiest to use slope-intercept form, which is y=mx + b, where m is the slope and b is the y-intercept.

We can plug in 2/3 for the slope, because we are given that.

y=2/3x + b.

To find the y-intercept, plug in the given point for x and y and solve for b.

-5 = 2/3 · (-3) + b

-5 = -2 + b

b = -3

So now we have the complete equation of the line.

<u>y = 2/3x -3</u>

Now to graphing.

We can put the y-intercept, (0, -3), and the given point, (-3, -5) on a graph.

3 0

3 years ago

Can someone find the sum??

harkovskaia [24]

Ok so the answer should be answer choice D.

Work: by combining all the like terms, the sum would be Answer choice D.

Hope this helps!

3 0

4 years ago

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It costs a general admission fee of $6 to get into the fair and then $1 for each ride.

Wewaii [24]

6$+ 1$ for every ride you want to ride.

8 0

4 years ago

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

3 years ago