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2 years ago

12

dente has been putting one penny in his bank everyday. now je has 161 pennies. for how many weeks has dente been adding pennies

to his bank?

2 answers:

aalyn [17]2 years ago

8 0

23

1 day is equal to 0.14285714 weeks, so multiply that by the number of days.

SIZIF [17.4K]2 years ago

7 0

23

1 day is equal to 0.14285714 weeks, so multiply that by the number of days.

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A city water company charges homeowners based on how much water they use in thousands of gallons. The company progressively char

Dima020 [189]

Answer:

1700: $50
2000: $60
7000: $90

Step-by-step explanation:

You can read the answers from the graph. Find the usage on the horizontal scale, and look at the vertical scale to find the corresponding charge.

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<h3>1700 gallons</h3>

The value 1700 is between 1000 and 2000, so will be between 1 and 2 on the horizontal scale. The graph is flat in that region. The charge is $50.

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<h3>2000 gallons</h3>

The graph has a jump at 2 on the horizontal scale. The solid dot tells you that the charge is $60.

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<h3>7000 gallons</h3>

At 7 on the horizontal scale, the graph is increasing from 80 to 100. At 7000 gallons, the line passes through the grid intersection indicating the charge is $90.

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1 year ago

JUST THE FIRST ONE EASY PLS ASAP

Sladkaya [172]

Answer:

<h2>For the first triangle, <u>x equals 22cm.</u></h2>

Here's how:-

x is the altitude or height of the triangle. To find it, we can use the Pythagoras theorem:-

$altitude^{2} + base^{2} = hypotenuse^{2}$

The base is 120cm and the hypotenuse is 122cm

Putting this in the equation, we get:-

$altitude^{2} + 120^{2} = 122^{2}$

$altitude^{2} = 122^{2}- 120^{2}$

$altitude^{2} = 14884 - 14400$

$altitude^{2}$ = 484

altitude = $\sqrt{484}$

<h3>altitude = 22cm</h3><h2><u>So, x = 22cm</u></h2>

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2 years ago

Aaaaaaaaaaaaaaaaaaaaa

tigry1 [53]

Step-by-step explanation:

what's the question about

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1 year ago

I NEED HELP PLEASE, THANKS! :)

NISA [10]

Answer: $\bold{b)\quad \dfrac{2\pi}{3},\dfrac{4\pi}{3},0}$

<u>Step-by-step explanation:</u>

Note the following identities: tan² x = sec²x - 1

$\sec x=\dfrac{1}{\cos}$

tan² x + sec x = 1

(sec² x -1) + sec x = 1

sec² x + sec x - 2 = 0

(sec x + 2)(sec x - 1) = 0

sec x + 2 = 0 sec x - 1 = 0

sec x = -2 sec x = 1

$\dfrac{1}{\cos x}=-2\qquad \dfrac{1}{\cos x}=1\\\\\cos x=-\dfrac{1}{2}\qquad \cos x=1\\\\x=\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\qquad x=0$

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2 years ago

Read 2 more answers

(Please show all work)

Lunna [17]

The answers here are

2) x= 2, 7

3) x= -5

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2 years ago