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Licemer1 [7]
3 years ago
11

Craig’s June water bill listed two meter reading. The previous reading was 6,372 ccf and the present reading is 6501 CFC. How mu

ch water did craigs household use during the billing period?
Mathematics
1 answer:
romanna [79]3 years ago
8 0

Answer: Craig's household use of water during the billing period = 129 CFC

Explanation:

Since we have given that

Craig's June water meter's previous reading =6372 CFC

Craig's water meter's present reading =6501 CFC

Craig's household use during the billing period =6501 -6372= 129 CFC

So, we get that

Craig's household use of water during the billing period = 129 CFC


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Y=2x-4 <br> What is the five ordered pairs and completing the input/output table
kati45 [8]

Answer:

{(0, -4), (1, -2), (2, 0), (3, 2), (4, 4)}

<u>Input values (x)</u> = {0, 1, 2, 3, 4}

<u>Output values (y)</u> = {-4, -2, 0, 2, 4}

Step-by-step explanation:

Given the linear equation, y = 2x - 4, you could practically choose any input value (for x), to have an output value (for y).

If x = 0:

y = 2(0) - 4

y = 0 - 4

y = -4

First ordered pair: y-intercept =  (0, -4)

If x = 1:

y = 2(1) - 4

y= 2 - 4

y = -2

Second ordered pair: (1, -2).

If x = 2:

y = 2(2) - 4

y= 4 - 4

y = 0

Third ordered pair: x-intercept =  (2, 0).

If x = 3:

y = 2(3) - 4

y = 6 - 4

y = 2

Fourth ordered pair: (3, 2).

If x = 4:

y = 2(4) - 4

y = 8 - 4

y = 4

Fifth ordered pair: (4, 4).

Your input values (x) = {0, 1, 2, 3, 4}

Your output values (y) = {-4, -2, 0, 2, 4}

6 0
3 years ago
Calculate the frequency in hertz of photons of light with energy of 8.10 × 10-19 J.
stira [4]
The Planck–Einstein relation E=hv gives that E, the photon energy, is equal to h, Planck's constant, times <span>ν</span>, the frequency.

We can solve the equation in terms of frequency to get
v= \dfrac{E}{h}

We substitute our given energy, 8.10 \times 10^{-19} J, and Planck's constant, 6.626 \times 10^{-34} Js.

v= \dfrac{8.10 \times 10^{-19} J}{6.626 \times 10^{-34} Js}=1.22 \times 10^{15} s^{-1}=1.22 \times 10^{15} Hz
5 0
3 years ago
Raymond has a set of 57 toy cars and decides to sell 19 of them to Joey. About how many toy cars does Raymond have left
Dovator [93]

Answer:

raymond has about 38 cars left

Step-by-step explanation:

Because raymond started off with 57 cars and he got rid of( take away) 19 you must use subtraction. 57 - 19 =38

4 0
3 years ago
Find the area of the composite figure.
irga5000 [103]

Answer:

Option C, 488 square centimeters

Step-by-step explanation:

<u>Step 1:  Determine the area of the rectangle</u>

A = l * w

A = 22\ cm * 17\ cm

A = 374\ cm^2

<u>Step 2:  Determine the area of the triangle</u>

A=\frac{h*b}{2}

A=\frac{19\ cm * (22\ cm-10\ cm)}{2}

A=\frac{19\ cm * (12\ cm)}{2}

A=\frac{228\ cm^2}{2}

A=114\ cm^2

<u>Step 3:  Determine the total area</u>

A_{total} = 374\ cm^2 + 114\ cm^2

A_{total} =488\ cm^2

Answer:  Option C, 488 square centimeters

4 0
2 years ago
Read 2 more answers
Tony took a city bus from his dorm to the school library, and then to a gym for a workout
Feliz [49]

Answer:

27 miles.

Step-by-step explanation:

Here I attach the draw of the coordinates.

Tony traveled 3 segments. The first was from (12,6) to (12, 15), where, leting 12 constant, he moved from 6 to 15 in the ordinates axis, which implies 9 units. This is the section 1 in the draw.

Then he moved from point B to C. If you notice, this distance is the hypotenuse on the the triangle DBC. We can find this value using Pitagoras' theorem:

DB^2 + CD^2 = CB^2

With DB=15 and CD=8 (12 minus 4 = 8)

15^2 + 8^2 = 289

So CB^2=289

Applying sqr root:

CB = 17

So, the second section has a measure of 17 units.

Finally, the 3rd section is the hypotenuse of the DAC triangle and we can use Pitagoras to solve it:

CD^2 + AD^2 = CA^2

8^2 + 6^2 = CA^2

64 + 36 = 100

So, CA=10

In the 3r section we traveled 10 units.

So, in total he traveled 10 + 17 + 9 = 36 units

As every unit is 0.75 miles he traveled 36*0.75 miles:

36*0.75 = 27 miles

He traveled in total 27 miles!!

6 0
3 years ago
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