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3 years ago

How many hours would you have to ride your bike at 10 mi/h to burn 550 calories? (Per Minute)

Mathematics

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

The time spend to burn energy is 3300 minutes .

Step-by-step explanation:

Given as :

The speed at which the bike ride = 10 miles per hours

The energy burn while riding = 550 calories

Let The time spend to burn energy = t minutes

So, According to question

Time spend = $\dfrac{\textrm Energy burn}{\textrm rate of ride}$

i.e t = $\dfrac{550}{10}$ hours

∴ t = 55 hours

So, in minutes

The time spend

1 hours = 60 min

∴ 55 hours = 60 × 55 = 3300 minutes

So, The spend to burn energy = t = 3300 minutes

Hence The time spend to burn energy is 3300 minutes . Answer

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When 24 is subtracted from a number and that difference is multiplied by 1/4, the result is 4.

Katen [24]

If (x-24)0.25 = 4, distribute 0.25 within the parenthesis.
0.25x - 6 = 4
0.25x = 10
Divide both sides by 0.25
x = 40

8 0

3 years ago

Mean of four numbers a ,b, c, and d is 8 and mean of seven

Nana76 [90]

Answer:

17 1/3 or 17.33

Step-by-step explanation:

Mean of four numbers a, b, c, and d is 8:

(a + b + c + d)/4 = 8
⇒ a + b + c + d = 32

Mean of seven numbers x, y, z, a, b, c and d is 12:

(x + y + z + a + b + c + d)/7 = 12

Substitute the sum of 4 numbers:

x + y + z + 32 = 7*12
x + y + z = 84 - 32
x + y + z = 52

Mean of x, y and z is:

52/3 = 17 1/3 or 17.33

4 0

2 years ago

The booster club at school is raising funds by selling t-shirts at each athletic event. They must pay the manufacturer $110 plus

insens350 [35]

Step-by-step explanation:

Total money paid=$110+$5.5=115.5

Cost of each shirt=$8

Therefore,

No. of T shirt req.=115.5/8=14.4

So they req. 15 T-Shirts

6 0

2 years ago

100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

4 0

3 years ago

Read 2 more answers

What is the vertex form of the parabola whose standard form equation is y=5x^2-30x+49

agasfer [191]

Answer:

The vertex is (3,4)

Step-by-step explanation:

To convert a quadratic from

y = a x 2 + b x + c

form to vertex form,

y = a ( x − h ) 2 + k , you use the process of completing the square.

First, we must isolate the x

terms:

y − 49 = 5 x 2 − 30 x + 49 − 49

y − 49 = 5 x 2 − 30 x

We need a leading coefficient of 1

for completing the square, so factor out the current leading coefficient of 2.

y − 49 = 5 ( x 2 − 6 x )

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by

on the left side of the equation. This is the coefficient we factored out in the previous step.

y − 49 + ( 5 ⋅ ? ) = 5 ( x 2 − 6 x + ? )

<- Hint:

62 = 3 ; 3 ⋅ 3 = 9

y − 49 + ( 5 ⋅ 9 ) = 5 ( x 2 − 6 x + 9 )

y− 49 + 45 = 5 ( x 2 − 6 x + 9 )

y − 4 = 5 ( x 2 − 6 x + 9 )

Then, we need to create the square on the right hand side of the equation:

y − 4 = 5 ( x − 3 ) 2

Now, isolate the y term:

y − 4 + 4 = 5 ( x − 3 ) 2 + 4

y − 0 = 5 ( x − 3 ) 2 + 4

The vertex is:

( 3 , 4 )

7 0

3 years ago