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Alchen [17]
3 years ago
15

Brad spend 3/5 of an hour every day doing his homework. How many hours did he spend on homework in 5 days.

Mathematics
1 answer:
belka [17]3 years ago
6 0

Answer: 3 hours spent on homework in 5 days

Step-by-step explanation:

3/5 of an hour is 36 min.

Multiply 36 by the number of days he spent doing homework (5)

36x5=180 and 180 is equivalent to 3 hrs.

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Can you help me with my homework please
umka21 [38]

#4(a)

row-seat: 3- 13, 4-15, 5-17, 6-19, 7-21

(b)

the equation works for row 1 but not for any of the rows after this

Ex: Row 2, S=7(2)+2, this would equal 14 but there isn't 14 seats in row #2

(c)

S=2(1)+7, there is 9 seats in row 1

2(2)+7=11,  there is 11 seats in row 2

2(3)+7= 13, there is 13 seats in row 3

(d)

2(15)+7=37

2*15=30, 30+7=37

(e)

91=2(r)+7

91-7=2(r)+7-7

84=2(r)

84/2=2(r)/2

42=r, the row with 91 seats is row 42


3 0
3 years ago
Simplify 6(x + 3).<br><br> 6x + 3<br> x + 18<br> 6x + 18<br> 6x + 9
Kisachek [45]

Answer:

the answer of this question is = 6x + 18

7 0
2 years ago
Is it possible to have the same lower quartile as the minimum in a box plot?
Natalka [10]
No, bc the lower quartile should be in between the minimum and middle quartile(median) Hoped this helped!
6 0
3 years ago
Read 2 more answers
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0&lt;×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
PLEASE HELP ASAPP!!!!!!!!!!!!!!!
max2010maxim [7]
\dfrac{1}{3^{-2}x^{-4}y^2}=\dfrac{3^2x^4}{y^2}\\\\\\&#10;\dfrac{3^2\cdot3^4}{(-1)^2}=\dfrac{3^6}{1}=3^6=\boxed{729}

Answer B)
3 0
3 years ago
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