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Dvinal [7]
3 years ago
9

Kelly bought a book of 20 stamps. She has used 80% of them. A. How many stamps has Kelly used? B. How many stamps does she have

left?
Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0
Hello there the answer is 110
jok3333 [9.3K]3 years ago
4 0
80% of 20 is 80/100 x 20 = 16

So, kelly has used 16 stamps, and has 4 left
You might be interested in
Choose all of the equivalent ratios for 64:80<br> 8:12<br> 32:40<br> 4:5<br> 16:24<br> 120:160
Alinara [238K]

Answer:

32/40, 4/5

Step-by-step explanation:

All these ratios are follow the ratio 4:5 meaning they are equivalent to "64:80".

8:12=0.666667

16"24=0.666667

120:160=0.75

5 0
3 years ago
Read 2 more answers
2,592+3,385 =3,385 + whar is it
amm1812
Lol... I don't know if this is a joke, but it's basically just restating the precious problem, but switching numbers. So the answer would be 2,592.
7 0
3 years ago
Tell whether (5. 10) is a solution of y = 6x – 18
almond37 [142]

Answer:

NO

Step-by-step explanation:

Simply enter 5, 10 in the x and y boxes to get the answer.

10=6(5)-18

10=30-18

10=12

As a consequence, it isn't a feasible option.

8 0
3 years ago
Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 9t + 9 cot(t/2), [π/4, 7π/4]
agasfer [191]

Answer:

the absolute maximum value is 89.96 and

the absolute minimum value is 23.173

Step-by-step explanation:

Here we have cotangent given by the following relation;

cot \theta =\frac{1 }{tan \theta} so that the expression becomes

f(t) = 9t +9/tan(t/2)

Therefore, to look for the point of local extremum, we differentiate, the expression as follows;

f'(t) = \frac{\mathrm{d} \left (9t +9/tan(t/2)  \right )}{\mathrm{d} t} = \frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}

Equating to 0 and solving gives

\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0

t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}

Where n_i is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively

Or we have by chain rule

f'(t) = 9 -(9/2)csc²(t/2)

Equating to zero gives

9 -(9/2)csc²(t/2) = 0

csc²(t/2)  = 2

csc(t/2) = ±√2

The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or

in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2

We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;

f(x) for x = π/4, π/2, 3π/2, 7π/2

f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965

f(π/2) = 9·π/2 +9/tan(π/4) = 23.137

f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412

f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96

Therefore the absolute maximum value = 89.96 and

the absolute minimum value = 23.173.

7 0
3 years ago
HELPPPPP PLESSEEE THIS IS FOR MY EXAMMM
Gelneren [198K]

Answer:

I think it is 80 12

Step-by-step explanation:

7 0
3 years ago
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