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2 years ago

Properties of equality

Mathematics

1 answer:

Inessa [10]2 years ago

5 0

1.D
2.E
3.C
4.G
5.A
6.E
7.B

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I need help answering these questions, methods appreciated on how to do each one.

xxMikexx [17]

Answer:

get rickrolled

Step-by-step explanation:

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2 years ago

Umm yah i know I suck at math

nasty-shy [4]

Its D because look if u add the white and red ones its 14 and all them together is 26 so it would be 14/26

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3 years ago

For her phone service, Rita pays a monthly fee of $17, and she pays an additional $0.05 per minute of use. The least she has bee

Alona [7]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Monthly fee = $17

Additional fee per unit of use = $0.05

Least amount she's been charged in a month = $83.90

Equation to represent the number of minutes 'M':

monthly fee + (additional fee per minute × M) = $83.90

Possible number of minutes :

83.90 ≤ $17 + 0.05M

$17 + $0.05 × M = $83.90

$0.05*M = $83.90 - $17

$0.05*M = $66.90

M = $66.90 / 0.05

M = 1,338 minutes

3 0

3 years ago

How can we solve fractions

tatuchka [14]

Answer:You can solve fraction by taking LCM

Step-by-step explanation:

For example: (1/4)+(1/2)is the question

Answer:LCM=4

(1/4)+(1*2/2*2)

(1/4)+(2/4)

(1+2/4)

(3/4)

Here is the answer

8 0

3 years ago

A camera is mounted at a point 4000 feet away from a geyser. If the water is rising vertically at 900 ft/s when it is 3000 feet

Leno4ka [110]

Answer:

$\dot \theta = 0.144\,\frac{rad}{s}$ , $\dot \theta = 8.251\,\frac{deg}{s}$ (Option B)

Step-by-step explanation:

The trigonometric diagram is included herein as attachment. The expression is presented below:

$\tan \theta = \frac{y}{x}$

Where:

$x$ - Horizontal distance between the geyser and the camera.

$y$ - Vertical distance between the geyser and the camera.

The rate of change in terms of time is:

$\dot \theta \cdot \sec^{2}\theta = \frac{\dot y\cdot x-y\cdot \dot x}{x^{2}}$

$\dot \theta \cdot \left(\frac{1}{\cos^{2}\theta} \right) = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}}$

$\dot \theta = \left(\frac{\dot y \cdot x - y \cdot \dot x}{x^{2}} \right)\cdot \cos^{2}\theta$

$\dot \theta = \left(\frac{\dot y \cdot x - y\cdot \dot x}{x^{2}} \right)\cdot \left(\frac{x^{2}}{x^{2}+y^{2}} \right)$

$\dot \theta = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}+y^{2}}$

Finally,

$\dot \theta = \frac{\left(900\,\frac{ft}{s} \right)\cdot (4000\,ft)-(3000\,ft)\cdot \left(0\,\frac{ft}{s} \right)}{(4000\,ft)^{2}+(3000\,ft)^{2}}$

$\dot \theta = 0.144\,\frac{rad}{s}$

$\dot \theta = 8.251\,\frac{deg}{s}$

3 0

3 years ago