I have 30 ones , 82 thousands ,4hundred thousands,60 tens, and 100hundreds what number am i

A dog walking service offers a daily and a weekly service. In the winter, the ratio of daily customers to weekly customers was 1

vazorg [7]

In winter:

The ratio of daily customers to weekly customers = 1:5.

In Spring :

The ratio of daily customers to weekly customers = 4:2.

Number of weekly customer in the winter are 125.

<em>Number of weekly customers are 5 times than the number of daily customers in winter.</em>

<em>Therefore, number of daily customers in winter = 125/5 = 25 customers.</em>

<h3>Number of daily customers were there in the spring = 4 × 25 = 100 customers.</h3>

5 0

3 years ago

azul has 4 green picks and no orange picks. you add only orange picks so 2 in every 3 picks are orange. how many picks are there

pychu [463]

You have 4 green picks and, after adding some orange picks, you know that every 2 picks from 3 are orange.

So, 1 green pick needs 2 orange picks, three in total. Since you have 4 green picks, then you will have in total 3·4=12 picks (4 green and 12-4=8 orange).

Answer: there are 12 picks in total.

5 0

3 years ago

Read 2 more answers

You are making identical gift bags using 12 crayons and 28 bottles of bubbles. What is the greatest number of gift bags you can

const2013 [10]

Since you don't want any leftover items, you have to find the greatest common factor (GCF) of both the crayons and bubbles. Since the highest common factor for both 28 and 12 is 4, the greatest number of gift bags you can make without any leftover items would be 4 gift bags.

Hope I could help.

4 0

3 years ago

Karen set her watch 1 second behind, and it falls behind another 2 seconds every day. How many days has it been since Karen last

Kay [80]

Answer:

The equation to find the answer is $9=1+(n-1)2$ .

It has been 5 days since Karen has set her watch.

Step-by-step explanation:

Given:

Karen set her watch 1 second behind

So we will consider;

First term 'a' = 1 second

Also Given :

it falls behind another 2 seconds every day.

So we can consider;

Common difference = 2

We need to find number of days it has been 9 seconds since Karen last set her watch

Solution:

Let number of days be represented by 'n'.

Also Let $T(n) = 9$

Now By Applying formula for Arithmetic Progression we get;

$T(n)=a+(n-1)d$

Substituting the given values we get;

$9=1+(n-1)2$

Hence The equation to find the answer is $9=1+(n-1)2$ .

On Solving the above equation we get;

Subtracting both side by 1 we get;

$9-1=1+(n-1)2-1\\\\8=(n-1)2$

Now Dividing both side by 2 we get;

$\frac{8}{2}=\frac{(n-1)2}{2}\\\\4=n-1$

Adding 1 on both side we get;

$4+1=n-1+1\\\\n=5$

Hence It has been 5 days since Karen has set her watch.

3 0

4 years ago

How do I solve this?Help me please!!!

NISA [10]

Given:
Ship M travels E 15 km, then N35E 27 km. Its sub travels down 48° 2 km from that location.

Ship F travels S75E 20 km, then N25E 38 km. The treasure is expected to be at this location 2.18° below horizontal from the port.

Find:
1a. The distance from port to Ship M
1b. The distance from port to the sub
1c. The angle below horizontal from the port to the sub

2a. The distance from port to Ship F
2b. The depth to the expected treasure location
2c. The distance from port to the expected treasure location

Solution:
It can be helpful to draw diagrams. See the attached. The diagram for depth is not to scale.

There are several ways this problem can be worked. A calculator that handles vectors (as many graphing calculators do) can make short work of it. Here, we will use the Law of Cosines and the definitions of Tangent and Cosine.

Part 1
1a. We are given sides 15 and 27 of a triangle and the included angle of 125°. Then the distance (m) from the port to the ship is given by the Law of Cosines as
m² = 15² +27² -2·15·27·cos(125°) ≈ 1418.60
m ≈ 37.66
The distance from port to Ship M is 37.66 km.

1b. The distance just calculated is one side of a new triangle with other side 2 km and included angle of 132°. Then the distance from port to sub (s) is given by the Law of Cosines as
s² = 1418.60 +2² -2·37.66·2·cos(132°) ≈ 1523.41
s ≈ 39.03
The distance from port to the sub is 39.03 km.

1c. The Law of Sines can be used to find the angle of depression (α) from the port. That angle is opposite the side of length 2 in the triangle of 1b. The 39.03 km side is opposite the angle of 132°. So, we have the relation
sin(α)/2 = sin(132°)/39.03
α = arcsin(2·sin(132°)/39.03) ≈ 2.18°
The angle below horizontal from the port to the sub is 2.18°.

Part 2
2a. We are given sides 20 and 38 of a triangle and the included angle of 100°. Then the distance (f) from the port to the ship is given by the Law of Cosines as
f² = 20² +38² -2·20·38·cos(100°) ≈ 2107.95
f ≈ 45.91
The distance from port to Ship F is 45.91 km.

2b. The expected treasure location is at a depth that is 2.18° below the horizontal from the port. The tangent ratio for an angle is the ratio of the opposite side (depth) to the adjacent side (distance from F to port), so we have
tan(2.18°) = depth/45.91
depth = 45.91·tan(2.18°) ≈ 1.748
The depth to the expected treasure location is 1.748 km.

2c. The distance from port to the expected treasure location is the hypotenuse of a right triangle. The cosine ratio for an angle is the ratio of the adjacent side to the hypotenuse, so we have
cos(2.18°) = (port to F distance)/(port to treasure distance)
(port to treasure distance) = 45.91 km/cos(2.18°) ≈ 45.95
The distance from the port to the expected treasure is 45.95 km.

Part 3
It seems the Mach 5 Mimi is the ship most likely to have found the treasure. That one seems ripe for attack. Its crew goes to a location that is 2.18° below horizontal. The crew of the FTFF don't have any idea where they are going. (Of course, the pirate ship would have no way of knowing if it is only observing surface behavior.)

5 0

3 years ago