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3 years ago

7

The rabbit population in a certain area is 500% of last years population. There are 1,300 rabbits this year. How many were there

last year?

1 answer:

kiruha [24]3 years ago

3 0

Answer:

The population last year was 260 rabbits

Step-by-step explanation:

1. multiply 100 by 1300

2. divide the answer by 500

3. that is the answer to your question

You might be interested in

What is the reduced answer to 4/9 multiples by 4/5 equal

deff fn [24]

First, we know that when multiplying fractions, we multiply both the numerator and denominator.

so, in 4/9 • 4/5,
4•4 = 16, and
9•5 = 45

so, 4/9 • 4/5 = 16/45.
now, we’ll look for the Least Common Factor

factors are numbers that you can multiply together to = another number.
the LEAST common Factor is the # that is smallest that you can divide both numbers by, in an equation and get a whole number.

for instance, 3•3 and 1•9 are the only ways to get 9, so, the factors are 1, 3, 9

let’s look for the LCF in 16 and 45. -
if we find the ways to get 16, we have:
1•16, 2•8, and 4•4
so, the factors are 1, 2, 4, 8, and 16.
this is called FACTORING :)

the ways to get 45 are...
1•45, 3•15, and 5•9, so the FACTORS are
1, 3, 5, 9, 15, & 45.
- compare the factors of 16 & 45,
none of them are the same besides 1, and we know that dividing these numbers by 1 will not do anything.

because of this, we can not reduce 16/45, so the reduced answer to 4/9 • 4/5 = 16/45

4 0

4 years ago

Find the value of the variable such that: the value of 5−a is 20 greater than the value of 6a−1

vovangra [49]

Answer:

-2

Step-by-step explanation:

(5 - a) = 20 + (6a -1) . . . . . 5-a is 20 greater than 6a-1

5 = 7a +19 . . . . . . . . . . . . . add a, collect terms

-14 = 7a . . . . . . . . . . . . . . . subtract 19

-2 = a . . . . . . . . . . . . . . . . divide by 7

The value of the variable is -2.

_____

<u>Check</u>

5 - (-2) = 7

6(-2) -1 = -13

7 is 20 greater than -13, so the answer checks OK.

4 0

4 years ago

A plane cuts a pyramid as shown in the diagram. What is the shape of the cross section

Andreas93 [3]

A is your answer! Because of how many sides it has.

6 0

3 years ago

Sarah had a bag of 16 marbles that are green and yellow. The number of yellow marbles is 2 less than double the number of green

Alex777 [14]

Answer:

10 yellow marbles

Step-by-step explanation:

(y is the amount of yellow marbles)

y=2x-2

y=2(6)-2

y=12-2

y=10

8 0

4 years ago

Read 2 more answers

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

a.

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

or

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$

Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.

We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,

$3 = 15-6*2$

Because 4 multiplies 3 to give 12, we multiply by 4

$12 = 15*4-6*8$

So one solution is

$x=-8$ & $y = 4$

All other solutions will have the form

$x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r$

where $r$ ∈ Ζ

Hence by putting r values, we get many (x, y)

3 0

3 years ago