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2 years ago

13

Can Someone please help me?!

1 answer:

valina [46]2 years ago

7 0

Answer:

The answer in exact form is 40/9, decimal form is 4.444..., and in mixed number form is 4 4/9.

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You forget the last two digits of your password for a website. What is the probability that you randomly choose the correct digi

Katen [24]

There are 100 possibilities for the last two digits [ if we include 00].....therefore....you have 1/100 chance [1 % ] chance of choosing the correct one.....

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3 years ago

What is m∠CDJ ? Enter your answer in the box. ° Line A K and line M J intersect at a right angle at point D. Line L C also passe

Wittaler [7]

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3 years ago

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Jesus, is it, a, b, c, or d??

serg [7]

Answer: A is the answer!

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3 years ago

A scientist mixes water (containing no salt) with a solution that contains 35% salt. She wants to obtain 245 ounces of a mixture

sesenic [268]

Answer:

140 ounces of water and 105 ounces of 35% salt solution.

Explanation:

Let $x$ be the ounces of water that need to be added and $y$ be the ounces of 35% solution that need to to be added.

We know that the total of ounces of water and of 35% solution must be 245 ounces:

$x+y=245$

35% of $x$ is the salt, and since no salt is added with addition of $y$ liters of water, this must be equal to the amount of salt in the final 15% solution:

$0.35x=0.15*245$

we solve for $x$ and get:

$x=105$

We put this value into $x+y=245$ and solve for $y$ to get:

$y=140$

Thus we have 140 ounces of water and 105 ounces of 35% solution.

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3 years ago

From a height of 50 meters above sea level on a cliff, two ships are sighted due west. The angles of depression are 61° and 28°.

Olenka [21]

Answer:

The ships are 66 meters apart.

Step-by-step explanation:

For the sake of convenience, let us label ships A and B

As shown in the figure, the distances to the ships from right triangles.

The distance to the ship A is $d_1$ and it is given by

$tan (61^o)= \dfrac{50}{d_1}$

$d_1=\dfrac{50}{tan (61^o)}$

$\boxed{d_1= 27.71m}$

And the distance to the ship B is $d_2$ and is given by

$tan (28^o)= \dfrac{50}{d_2}$

$d_2=\dfrac{50}{tan (28^o)}$

$\boxed{ d_2=94.04m}$

Therefore, the distance $d$ between the ships A and B is

$d= d_2-d_1=94.04-27.7\\\\\boxed{d=66m}$

In other words, the ships are 66 meters apart.

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3 years ago

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