All categories

2 years ago

A horse ate 1.8 pounds of grass in 4 days how many grass can the horse eat for 7 days

Mathematics

2 answers:

aleksandr82 [10.1K]2 years ago

7 0

Answer:

3.15 pounds of grass.

Step-by-step explanation:

If you take the amount of pounds the horse ate and divide it by how many days it took then you get 0.45.

0.45 x 7 = 3.15

jolli1 [7]2 years ago

4 0

i think the answer to this is 3.15

You might be interested in

4x2 minus 20+48 factorise this polynomial completely

bulgar [2K]

Answer:

Step-by-step explanation:

With no parenthesis

7 0

3 years ago

Read 2 more answers

I may be like 12 weeks in, but I still need help with coefficients and what they are. (Btw my state does NOT use common core met

STatiana [176]

Answer:

They are any number in front of a variable so like 4x 7y or 8z. 4, 7 and 8 are the coefficients. A number by itself is a constant. For example 3, 1 and 9 are constants.

6 0

2 years ago

Read 2 more answers

PRE-CALC<br><br>Determine when 2x^3 + 3x^2 – 17x - 30 < 0 using a sign chart.

Oksi-84 [34.3K]

Step-by-step explanation:

Use the Rational Zero Theorem to list all possible rational zeros of the function.

Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. ...

Repeat step two using the quotient found with synthetic division. ...

Find the zeros of the quadratic function.

3 0

3 years ago

# 27 a rectangular piece of cloth had an area of 1/12 square yard

Margaret [11]

1/9
becuase 27-1/12= 1/9

3 0

2 years ago

Read 2 more answers

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago