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mixer [17]
3 years ago
10

Decrease 60 kg by 25%​

Mathematics
2 answers:
nekit [7.7K]3 years ago
7 0

Answer:

30

Step-by-step explanation:

60×25/100

=30

Ans: 30

anygoal [31]3 years ago
6 0
100 - 25 = 75%
75% = 0.75
60 x 0.75 = 45kg
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A bakery makes 40 different flavors of muffins. 25% of the flavors have chocolate as one of the ingredients. How many flavors ha
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Answer:

10

Step-by-step explanation:

because 25%of 40 is 10

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3 years ago
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Given the literal equation for the area formula of a trapezoid: A =12h(b1 + b2) solve for b1. Must show your work to receive ful
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Use photo math it is a app

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3 years ago
Help plz:))) I’ll mark u brainliest <br> ASAP!!!
RSB [31]

Answer:30/4

Step-by-step explanation:180-11x+5, add all up=180, you get 30/4

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7 0
3 years ago
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An ant crawls at a rate of 2 1/6 ft per minute. At this rate, how long does it take the ant to crawl a distance of 5 1/5 ft? Ent
tangare [24]
For 2 1/6 = 13/6 Ft, ant takes = 1 min.
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Now, for 5 1/5 = 26/5 min. it would take = 6/13 * 26/512/5 = 2.4 min

In short, Your Answer would be 2.4 min

Hope this helps!
3 0
3 years ago
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Find lim h-&gt;0 f(9+h)-f(9)/h if f(x)=x^4 a. 23 b. -2916 c. 2916 d. 2925
Svetach [21]

\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = \lim_{h\to0}\frac{(9+h)^4-9^4}h

Carry out the binomial expansion in the numerator:

(9+h)^4 = 9^4+4\times9^3h+6\times9^2h^2+4\times9h^3+h^4

Then the 9⁴ terms cancel each other, so in the limit we have

\displaystyle \lim_{h\to0}\frac{4\times9^3h+6\times9^2h^2+4\times9h^3+h^4}h

Since <em>h</em> is approaching 0, that means <em>h</em> ≠ 0, so we can cancel the common factor of <em>h</em> in both numerator and denominator:

\displaystyle \lim_{h\to0}(4\times9^3+6\times9^2h+4\times9h^2+h^3)

Then when <em>h</em> converges to 0, each remaining term containing <em>h</em> goes to 0, leaving you with

\displaystyle\lim_{h\to0}\frac{f(9+h)-f(9)}h = 4\times9^3 = \boxed{2916}

or choice C.

Alternatively, you can recognize the given limit as the derivative of <em>f(x)</em> at <em>x</em> = 9:

f'(x) = \displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h \implies f'(9) = \lim_{h\to0}\frac{f(9+h)-f(9)}h

We have <em>f(x)</em> = <em>x</em> ⁴, so <em>f '(x)</em> = 4<em>x</em> ³, and evaluating this at <em>x</em> = 9 gives the same result, 2916.

8 0
3 years ago
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