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kati45 [8]
3 years ago
11

The number of fish in a lake decreased by 25% between last year and this year last year there were 60 fish in the lake what is t

he population this year? If you get stuck consider drawing a diagram
Mathematics
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

45

Step-by-step explanation:

Decreasing by 25% means that there is 25% less of last years amount or 0.25x less. It also means that this year there is only 75% of what was last year actually in the lake or 0.75x. Since last year, there were 60 fish, this means this year there is 0.75(60) = 45. This year has 45 fish.

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Solve t^4-2t^3+24t^2=0​
AlexFokin [52]

Answer:

t=0, 6, -4

Step-by-step explanation:

first, you can factor out t² of the equation

t²(t²-2t+24)=0, then factor the trinomial

t²(t-6)(t+4)=0, now set each factor equal to zero

t²=0, so t = 0

t-6=0, so t = 6

t+4=0, so t = -4

If you graph it, you can also see the answers.

4 0
4 years ago
Write a real world problem that involves a percent then model the percent used in the problem
s344n2d4d5 [400]
You go to a restaurant and your bill is $35.00. Your service was excellent so you leave a 20% tip. What was the total amount paid for the bill including the tip?

10%=$1 for every $10
15%=$1.50 for every $10
20%=2% for every $10

10% of the bill would be $3.50(move the decimal to the tens place)
15% of the bill would be $5.25( halve the 10% and add it to the original 10% tip; $3.50+$1.75)
20% of the bill would be $7.00(double the 10% tip)
7 0
4 years ago
Plz help I’ll give Brainly
MrRissso [65]

Answer:

[i] Area of Juan's garden:

Soln,

length(l)=16

width(w)=12

area(a)=?

Now,

a=l×w

=16×12

=192

[ii] Area of Matthew's garden:

Soln,

length(l)=18

width(w)=21

area(a)=?

Now,

a=l×w

=18×21

=378

So,the area ofJuan's garden=192

And

the area of Matthew's garden=378

7 0
3 years ago
"(b) use part (a) to find a power series for f(x) = 1 (4 + x)3 ."
Firdavs [7]
Assuming

f(x)=\dfrac1{(4+x)^3}

Recall that for |x|,

\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}

Denote the above by s(x). Then

s'(x)=\displaystyle\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}
s''(x)=\displaystyle\sum_{n\ge2}n(n-1)x^{n-2}=\frac2{(1-x)^3}

Now,

\dfrac1{(4+x)^3}=\dfrac1{4^3}\dfrac1{\left(1-\left(-\frac x4\right)\right)^3}=\dfrac1{2^7}\dfrac2{\left(1-\left(-\frac x4\right)\right)^3}

which means we have

f(x)=\dfrac1{2^7}s''(x)=\dfrac1{128}\displaystyle\sum_{n\ge2}n(n-1)\left(-\frac x4\right)^{n-2}

which is valid for \left|-\dfrac x4\right|, or |x|.
4 0
3 years ago
the work of johnannes kepler is likely to have been most influenced by the work of which of the following scientists
dedylja [7]
Kepler was a student of Copernicus who went with observations he made instead of tradition.
6 0
3 years ago
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