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d1i1m1o1n [39]
3 years ago
10

The length of a jogging trail is 3528 m. A jogger wants to complete the trail within 30 min. How many meters must the jogger tra

vel each minute?
Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0
<h3>Answer: dt= 117 (3/5) <em>m</em></h3>

Step-by-step explanation:

We have to find the distance the jogger must travel each minute. The distance covered in one minute is:

d= 3528/1800= 1.96 <em>m</em>

Therefore, the distance covered in one minute, or 60 seconds is:

dt= 1.96 × 60 = 117.6 <em>m</em>

The solution needed is, dt=588/5 = 117 (3/5) <em>m</em>

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On a map Rossville and Smithtown are 5 1/2 inches apart. If the actual distance between the town is 16 1/2 miles, to what unit r
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c

Step-by-step explanation:

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0.00000003 in scientific notation
tester [92]

Answer:

3×10x^{-8}

Step-by-step explanation:

0.00000003

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2 years ago
What is it called in cubic inch of a cylinder with a height of 4in and a base radius of 5in to the nearest tenths place?
Maru [420]

Answer:

1,256.8in

Step-by-step explanation:

Given data

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V=πr²h

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6 0
3 years ago
2. f(x)=4x+1 and ​g(x)=x2−4x−5​, find ​(g◦​f)(4​).
Goryan [66]

Answer:

(g · f)(4) = 45

Step-by-step explanation:

f(x)=4x+1

g(x)=x² - 4x- 5

(g · f)(x) = 4(x² - 5) + 1

(g · f)(4) = 4(4² - 5) + 1

Following pemdas

(g · f)(4) = 4(16 - 5) + 1

(g · f)(4) = 4(11) + 1

(g · f)(4) = 44 + 1

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8 0
1 year ago
Suppose that 5 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 39 cm.
vesna_86 [32]

Answer

given,

work = 5 J

spring stretch form 30 cm to 39 cm

W = \dfrac{1}{2}kx^2

x = 0.39 - 0.30 = 0.09 m

5 = \dfrac{1}{2}k\times 0.09^2

k = \dfrac{5\times 2}{0.09^2}

k = 1234.568 N/m

a) work when spring is stretched from  32 cm to 34 cm

x₂= 0.34 -0.30 = 0.04 m

x₁ = 0.32 - 0.30 = 0.02

W = \dfrac{1}{2}k(x_2^2-x_1^2)^2

W = \dfrac{1}{2}\times 1234.568 \times (0.04^2-0.02^2)^2

W = 0.741 J

b) F = k x

  25 = 1234.568 × x

     x = 0.0205 m

     x = 2.05 cm

3 0
3 years ago
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