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Ainat [17]
3 years ago
7

While lifting weights, Jacob adds 9.08 kilograms to his bar. About how many pounds does he add to his bar?

Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

20

Step-by-step explanation:

First you need to divided 9.08 by 0.454 which equals 20 and so that is your answer

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What is the value of x?
Ray Of Light [21]
The Pythagorean Theorem states that a^2+b^2=c^2
We already have:
A^2+6^2=117
All we need to do is find the missing A by subtract b from c.
117-36=81
Now the square root of 81, 9.
So, x is 9 cm.
6 0
3 years ago
Just give me the answer
Hoochie [10]

Answer:

1/32768

Step-by-step explanation:

I don't know how to explain it to you. I am sorry.

8 0
3 years ago
At 5.45p.M .,lixin,khairul and Devil are at the starting point of a 1-km circular path. Lixin Takes 18 minutes to walk one round
sweet [91]

Answer:

Meeting\ Time = 6:03\ pm

Step-by-step explanation:

Given

Start\ Time= 5:45pm

Lixin = 18\ min/round

Khairul = 360\ seconds/round

Devi = 4minutes/2\ rounds

Required

Determine their next meeting time

First, we need to get their unit rate in minutes per round

Lixin = 18\ min/round

Khairul = 360\ seconds/round

Convert 360 seconds to minutes

Khairul = \frac{360}{60} min/round

Khairul = 6 min/round

Devi = 4minutes/2\ rounds

Divide by 2/2

Devi = 2mins/round

So, we have:

Lixin = 18\ min/round

Khairul = 6 min/round

Devi = 2mins/round

Next, is to determine the LCM of 18, 6 and 2

18=2 * 3 * 3

6 = 2 * 3

3 = 3

Their LCM is:

LCM = 2 * 3 * 3

LCM = 18

This means that they will meet 18 minutes after their initial time of departure.

So:

Meeting\ Time = 5:45\ pm + 18\ minutes

Meeting\ Time = 6:03\ pm

3 0
3 years ago
If the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20
zysi [14]

Answer:

Therefore, we conclute that the map distance between A and C is 35 m.u..

Step-by-step explanation:

We know that the order of three loci is A B C, and the map distance between A and B is 15 m.u., and the map distance between B and C is 20 m.u. We calculate the map distance between A and C.

Therefore, we get

AC=AB+BC\\\\AC=15+20\\\\AC=35\\

Therefore, we conclute that the map distance between A and C is 35 m.u..

4 0
3 years ago
Urgent!! Will mark brainliest!!
horsena [70]

Answer:

1) x is negative and y is positive ⇒ last answer

2) cotФ = -12/35 ⇒ second answer

3) The right identity is cot²Ф - csc²Ф = -1 ⇒ last answer

Step-by-step explanation:

* For any point (x , y) lies on the terminal side of the angle Ф

 in standard position

* x = cosФ and y = sinФ

- If Ф in the first quadrant, then x , y are positive

∴ All trigonometry functions are positive

- If Ф in the second quadrant, then x is negative , y is positive

∴ sinФ only is positive

- If Ф in the third quadrant, then x is negative , y is negative

∴ tanФ only is positive

- If Ф in the fourth quadrant, then x is positive , y is negative

∴ cosФ only is positive

* Lets solve the problems

∵ Ф = 3π/4 ⇒ (135°)

∴ It lies on the second quadrant

∴ x is negative and y is positive

* Lets revise the reciprocal of sinФ, cosФ and tanФ

- cscФ = 1/sinФ

- secФ = 1/cosФ

- cotФ = 1/tanФ

∵ secФ = -37/12

∴ cosФ = -12/37

∵ π/2 < Ф < π

∴ Ф lies on the second quadrant

∴ cotФ is negative values

∵ tan²Ф = sec²Ф - 1

∵ secФ = -37/12

∴ tan²Ф = (-37/12)² - 1 = 1225/144 ⇒ take√ for both sides

∴ tanФ = ± 35/12

∵ cotФ = ± 12/35

∵ cotФ is negative value

∴ cotФ = -12/35

* In the standard position of the angle Ф the terminal

 of it lies on the unit circle O

- By using Pythagorean theorem

∵ x² + y² = 1

∵ x = cosФ and y = sinФ

∴ cos²Ф + sin²Ф = 1 ⇒ (1)

∴ cos²Ф = 1 - sin²Ф

∴ sin²Ф = 1 - cos²Ф

* Divide (1) by cos²Ф

∴ cos²Ф/cos²Ф + sin²Ф/cos²Ф = 1/cos²Ф

* Remember sin²Ф/cos²Ф = tan²Ф and 1/cos²Ф = sec²Ф

∴ 1 + tan²Ф = sec²Ф ⇒ (2) ⇒ subtract 1 from both sides

∴ tan²Ф = sec²Ф - 1 ⇒ subtract sec²Ф from both sides

∴ tan²Ф - sec²Ф = -1

* Divide (1) by sin²Ф

∴ cos²Ф/sin²Ф + sin²Ф/si²Ф = 1/sin²Ф

* Remember cos²Ф/sin²Ф = cot²Ф and 1/sin²Ф = csc²Ф

∴ cot²Ф + 1 = csc²Ф ⇒ (3) ⇒ subtract 1 from both sides

∴ cot²Ф = csc²Ф - 1 ⇒ subtract csc²Ф from both sides

∴ cot²Ф - csc²Ф = -1

* The right identity is cot²Ф - csc²Ф = -1

3 0
3 years ago
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