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Tomtit [17]
4 years ago
14

Keenan currently does a total of 8 pushups each day. He plans to increase the number of pushups he does each day by 2 pushups un

til he is doing a total of 30 pushups each day.
Mathematics
2 answers:
Viefleur [7K]4 years ago
5 0

Answer:

12

Step-by-step explanation:

Given that

Current count of pushups = 8

Number of increase in pushups each day = 2

The desired pushup goal on daily basis = 30

Days require = x= ?

as we can conclude the number of pushups would increase by following sequence

8, 10, 12, 14, 16, ......30

Now as we can be the number of pushups by a constant number each day it means this sequence is arithmetic.

First term of our sequence is X1 and the final term is Xn, and the difference between two consecutive terms (Common Difference) is 2.

So

X1 = 8

Xn = 30

d = 2

Using the formula of nth term for arithmetic sequence

Xn = X1 + (n-1)d

Substituting the values

30 = 8 + (n-1)2

30-8 = (n-1)2

22 = 2n -2

22+2 = 2n

2n = 24

n = 24/2

n = 12

So, it means on the 12th day he will have to do 30 pushups if he starts from 8 and increase 2 pushups each day.


Assoli18 [71]4 years ago
5 0

Answer:

8 + 2x = 30

Step-by-step explanation:

I did it on khan academy...

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PLS SOLVE ASAP!!!
cupoosta [38]
180 + 30x = 60 + 50x
⇒ 20x = 120
⇒ x = 6 minutes

<span>it will take 6 minutes for Millie to catch up with Ben</span>

3 0
3 years ago
What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

                                           = 5.2 feet

r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

                            = 3.14 x (2.6)^{2}

                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

                                           = 21.2264 - 19.625

                                           = 1.6014

The cross sectional area of the cement pipe is 1.60 ft^{2}.

6 0
3 years ago
Please help!!!!!!!! due today
lozanna [386]

Answer:

148⁰

Step-by-step explanation:

Since

we know that the angle in a straight line is 180⁰

<axb+<bxd=180⁰

Given

<axb=32⁰

Therefore

<bxd=180⁰-32⁰=148⁰

Option D

4 0
2 years ago
the Length, breadth and height of a wall is 5m,30cm and 3 m how many bricks of dimensions 20cm ×10 cm× 7.5 cm are required to bu
raketka [301]

Answer:

<h2>3000 pcs</h2>

Step-by-step explanation:

given:

the Length, breadth and height of a wall is 5m,30cm and 3 m

find:

how many bricks of dimensions 20cm ×10 cm× 7.5 cm are required to build the wall​

solution:

vol of the wall = 5 x 3 x 0.30= 4.5 m³

vol per brick = 0.20 x 0.10 x 0.075 = 0.0015 m³

number of bricks =  <u> wall volume </u>

                                 vol. per brick

                             = 4.5 / 0.0015

                             = 3000 pcs of bricks required

6 0
3 years ago
:
Scorpion4ik [409]

Answer:

  x-intercept: (4, 0)

  y-intercept: (0, 3)

Step-by-step explanation:

The intercepts of the equation can be found by setting each variable to zero and solving for the other. The graph will be a line through the intercept points.

<h3>Intercepts</h3>

Setting x=0 and solving for y, we have ...

  4y -12 = 0   ⇒   y = 12/4 = 3

Setting y=0 and solving for x, we have ...

  3x -12 = 0   ⇒   x = 12/3 = 4

The intercepts are ...

  x-intercept: (4, 0)

  y-intercept: (0, 3)

<h3>Graph</h3>

The line will go through these intercept points.

3 0
2 years ago
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