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olga55 [171]
3 years ago
11

What is the distance between the points (4, 7) and (4, −5) ?

Mathematics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

12 units

Step-by-step explanation:

the 2 points have the same x but different y so you would find the difference of the y values

|-5|+|7|=12

12units away

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Can someone please take a moment to draw me something like this for my project pleaseee only 1 is good (the on In the photo) and
Kay [80]

Answer:

i dont understand this question at all

Step-by-step explanation:

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3 years ago
Help with part a please
julsineya [31]
Use the sum of angles's trigonometric identity formula:
cos(A+B)=(cosAcosB-sinAsinB)
x+y=4cos(t+π/6)+2sint=4(cost*cosπ/6-sint*sinπ/6)+2sint
recall that cosπ//6=√3/2, and sinπ/6=1/2:
4(cost*cosπ/6-sint*sinπ/6)+2sint=4[(√3/2)cost-(1/2)sint]+2sint
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3 years ago
Solve the inequality: 5x + 20 > 50
Nostrana [21]

x=6

5x  + 20 > 50 \\ 5x > 50 - 20 \\ 5x > 30 \\  \frac{5x}{5}  >  \frac{30}{5 }  \\ x > 6

8 0
3 years ago
Kayla is on level 12 of a new game. Everyday she completes 6 new levels. Maddy is on level 28 of the same game. Everyday she com
ankoles [38]

Answer:

Kayla and Maddy will be in the same level in 4 days.

Step-by-step explanation:

Kayla: y = 6x + 12

Maddy: y = 2x  + 28

6x + 12 = 2x + 28

Subtract 2x from both sides;

4x + 12 = 28

Subtract 12 from both sides;

4x = 16

Divide both sides by 4;

x = 4

3 0
3 years ago
Read 2 more answers
A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its
VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate (\dot V), measured in liters per hour, is directly proportional to draining time (t), measured in hours. That is:

\dot V \propto \frac{1}{t}

\dot V = \frac{k}{t} (Eq. 1)

Where k is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

\dot V_{s}+\dot V_{l} = \frac{k}{2} (Eq. 2)

\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}

(ii) <em>Only the small drain is opened</em>

\dot V_{s} = \frac{k}{t_{l}+3} (Eq. 3)

\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}

(iii) <em>Only the big drain is opened</em>

\dot V_{l} = \frac{k}{t_{l}} (Eq. 4)

\frac{\dot V_{l}}{k}  = \frac{1}{t_{l}}

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}

\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}

2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}

t_{l}^{2}-t_{l}-3 = 0 (Eq. 5)

Whose roots are determined by the Quadratic Formula:

t_{l,1}\approx 2.303\,h and t_{l,2} \approx -1.302\,h

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

t_{s} = 2.303\,h+3\,h

t_{s} = 5.303\,h

The draining time when only the small drain is opened is 5.303 hours.

7 0
3 years ago
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