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IgorLugansk [536]
2 years ago
10

A puppy's weight increases by 15% each week for the first 6 months. The puppy currently weighs 3.7 pounds. Which expression will

not calculate how much the puppy will weigh in 1 week?
Mathematics
1 answer:
attashe74 [19]2 years ago
8 0

Answer:

y=3.7+0.15(1)

Step-by-step explanation:

<em>Note we are not given the options to choose from in this question, let us solve for the answer directly</em>

Step one:

given data

The puppy's weight increases by 15% each week for the first 6 months

=0.15

The puppy currently weighs 3.7 pounds

Step two:

let the number of weeks be x

and the weight after x weeks be y

so

y= 3.7+0.15x

after 1 week the expression to calculate the weight is

y=3.7+0.15(1)

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kap26 [50]

Answer:

15°

Step-by-step explanation:

AB is parallel to DC and AD is parallel to BC

Therefore, if you put Angle D and Angle C together it must form 180°

If Angle C = 120°, that means Angle D = 60°

Now you subtract 45 from 60 and that gives you 15.

3 0
2 years ago
This was the question it got deleted I just need 1,2,5
scoundrel [369]

Answer:

1. The volume of the cylinder is 263 cm^{3}.

2. The volume of the cylinder is 12 m^{3}.

3. The volume of the cylinder is 3165.1 cm^{3}.

4. The radius of the cylinder is 3.6 cm.

5a. Area of the base is 50.2 cm^{2}.

5b. The curved surface area of the pencil holder is 351.7 cm^{2}.

5c. The surface are of the pencil holder is 452.2 cm^{2}.

Step-by-step explanation:

1. The given shape is a cylinder, so that;

volume of a cylinder = \pir^{2}h

                                  = 3.14 x (3.8)^{2} x 5.8

                                  = 262.98128

The volume of the cylinder is 263 cm^{3}.

2. This is also a cylinder, but with a given diameter.

radius = \frac{diameter}{2}

          = \frac{1.2}{2} = 0.6 m

volume of a cylinder = \pir^{2}h

                                  = 3.14 x (0.6)^{2} x 10.6

                                  = 11.98224

The volume of the cylinder is 12 m^{3}.

3. volume of a cylinder = \pir^{2}h

                                      = 3.14 x 6^{2} x 28

                                      = 3165.12

The volume of the cylinder is 3165.1 cm^{3}.

4. volume of a cylinder = \pir^{2}h

239 = 3.14 x r^{2} x 6

239 = 18.84 r^{2}

r^{2} = \frac{239}{18.84}

   = 12.6858

r = \sqrt{12.6858}

 = 3.5617

The radius of the cylinder is 3.6 cm.

5. Area of the base = \pir^{2}

radius = \frac{diameter}{2}

           = \frac{8}{2} = 4 cm

Area of the base = 3.14 x 4^{2}

                            = 50.24

Area of the base is 50.2 cm^{2}.

5b. Area of the curved surface of the pencil holder = 2\pirh

                                                                    = 2 x 3.14 x 4 x 14

                                                                     = 351.68

The curved surface area of the pencil holder is 351.7 cm^{2}.

5c. Surface area of the pencil holder = 2\pirh + 2\pir^{2}

                               = 2\pir(h + r)

                                = 2 x 3.14 x 4 (14 + 4)

                                = 452.16

The surface are of the pencil holder is 452.2 cm^{2}.

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2 years ago
What is the slope of the line passing through the points (-3,-5) and (2,-5)
kogti [31]

Answer

m = 10

Step-by-step explanation:

Formula:

m = (y2 - y1) / (x2 - x1) => plug in points

(-3,-5) and (2,-5)

3 0
2 years ago
use symbols to compare the numbers 512 and 521. write the comparison to ways. then explain your thinking
Morgarella [4.7K]

Solution:-  Given numbers to compare are 512 and 521 .

As they are 3 digit numbers

So, we have to compare hundreds place.

But hundreds are equal in both the numbers with digit 5.

Next we have to compare tens place.

Case 1 :- 1 ten is smaller in 512 than 2 tens in 521 .

So we get the result that,

512 is smaller than 521

or <em>  512 < 521</em>

Case 2:-2 tens is greater in 521 than 1 ten in 512 .

So we get the result that,

521 is greater than 512

or  <em> 521 > 512</em>


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3 years ago
Simplify a - {b - [c - (d - e) - f] - g}.
GrogVix [38]

i hope it can help u!!!!!!!!!!

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3 years ago
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