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3 years ago

Please help it is due today

Mathematics

1 answer:

just olya [345]3 years ago

8 0

Answer:

4 2/5 tons

Step-by-step explanation:

2/5 = 4/10 = .40

4.40 tons > 4.3 tons

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3. use the law of exponents to evaluate (1/x) ^ -n when x= 7 and n= 2

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D on both ...................

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4 years ago

What is the sum of the first 10 terms of the sequence defined by an=3n-3?

netineya [11]

The answer is D. 135

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3 years ago

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Adriana va a llenar bolsas de dulces para su fiesta de cumpleanos si tiene un paquete con 100 paletas y otro con 56 chocolates y

lions [1.4K]

Answer:

56 bags of a chocolate and a popsicle

Step-by-step explanation:

The first thing is to find the relationship between the number of popsicles and the number of chocolates, which would be the quotient of these:

100/56 = 1.78

Since it does not reach number two, you cannot add 2 popsicles and 2 chocolates, therefore the only way that they contain the same amount is that there is a chocolate and a popsicle, that is to say that a total of 56 bags would come out.

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3 years ago

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

ira [324]

Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .

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3 years ago

Find the GCF 1. 12,21,30 2. 11,44

Dafna1 [17]

The GCF of 12, 21, and 30 is 3.
The factors of 12 are: 1, 2, 3, 4, 6, 12
The factors of 21 are: 1, 3, 7, 21
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
Then the greatest common factor is 3.

The GCF of 11, and 44 is 11.
The factors of 11 are: 1, 11
The factors of 44 are: 1, 2, 4, 11, 44
Then the greatest common factor is 11.

Hope this helps.

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3 years ago

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