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azamat
3 years ago
7

Joaquin goes out for a long walk. He walks 5/12 of a mile and then sits down to take a rest. Then he walks 1/3 of a mile. How fa

r did Joaquin walk altogether?
Mathematics
1 answer:
Vsevolod [243]3 years ago
6 0
The answert to ur question is 3/4
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-- 500 centimeters  =  5 meters

-- 0.005 kilometers  =  5 meters

-- 5,000,000,000 nanometers  =  5 meters
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Solve part a and part b
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Step-by-step explanation:

kusurs bakama yapamadım

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Write a quadratic equation in the form ax^2+bx+c=0 with roots of -4 and 5
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Step-by-step explanation:

- 4 & 5 are roots

Therefore, (x + 4) & (x - 5) are factors.

Required quadratic equation is given as:

(x + 4) (x - 5)  \\  = x(x - 5) + 4(x - 5) \\  =  {x }^{2}  - 5x + 4x - 20 \\  =  {x}^{2}  - x - 20 \\

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3 years ago
A man saves 4% of his monthly
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Answer:

Although the question is not clear, It most likely looks like you were asking for the calculation of the savings for the month after increase.

savings for the month after increase = $1172.4

Step-by-step explanation:

First, let us calculate how much was saved before the increase in savings:

monthly income = $19,540

Percentage saved = 4% of monthly income

= 4/100 × 19,540 = 0.04 × 19,540 = $781.6

Next, we are given the ratio of increase in savings as 3:2

Let the new savings amount be x

3 : 2 = x : 781.6

\frac{3}{2} = \frac{x}{781.6} \\781.6\ \times 3\ =2x\\2344.8 = 2x\\x =\frac{2344.8}{2} \\x = \$1172.4

therefore savings for the month after increase = $1172.4

Just incase you were looking for the savings before the increase, the answer is $781.6 (as calculated above)

4 0
3 years ago
Given that S^3 on the bottom 1 (e^x)dx=(e^3)-e use the properties of integrals and this result to evaluate S^3 on the bottom 1 (
scZoUnD [109]
Remember that
1) For functions f(x) and g(x), \int {f(x) + g(x)} \, dx =  \int {f(x)} \, dx + \int {g(x)} \, dx
2) For a function f(x) and a constant c, \int {cf(x)} \, dx =  c \int {f(x)} \, dx

Using these two properties of integrals, and the fact that \int\limits^3_1 {e^x} \, dx = e^3 - e, we can see that

\int\limits^3_1 {5e^x - 1} \, dx
= \int\limits^3_1 {5e^x} \, dx - \int\limits^3_1 1} \, dx
= 5 \int\limits^3_1 {e^x} \, dx - \int\limits^3_1 1} \, dx
= 5(e^3 - e) - \left.x\right|_1^3
= 5e^3 - 5e - (3 - 1)
= \bf 5e^3 - 5e - 2
6 0
3 years ago
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