All categories

3 years ago

How do you solve 2/5 (c+8.5) =18

Mathematics

1 answer:

Orlov [11]3 years ago

4 0

Solve for c by simplifying both sides of the equation, then isolating the variable.

C=36.5

You might be interested in

In m + 8, the variable m can have any value. Tell whether each statement is true or false. Then explsin your reasoning.

mote1985 [20]

M can have any value because M can represent any number because it is the unknown.

4 0

3 years ago

I will give extra points please help

Jlenok [28]

Answer:

x/3≤5

Step-by-step explanation:

Write the sentence into symbols

4 0

3 years ago

Math help, please! cannot do the last one

ryzh [129]

Answer:

-99.9

Step-by-step explanation:

7 0

3 years ago

The sum of three numbers is -2. The sum of three times the first number, twice the second number, and the third number is nine.

Anna35 [415]

Answer:

The numbers are 3, 5, -10.

Step-by-step explanation:

Let, the numbers are X, Y, Z.

Given, sum of the three numbers are -2.

So, the equation is X + Y+ Z = -2 .......(i)

Again, for 2nd condition, the equation will be

3X + 2Y + Z = 9 ....... (ii)

Again, for the third condition, the equation will be

Y - (Z/2) = 10 ......(iii)

Multiplying the equation (i) by 3 and the equation (ii) by 1, we get

3X + 3Y+ 3Z = -6

(Deduct) Y + 2Z = -15 ......(iv)

Solving equation (iii) and (iv) we get,

Y - (Z/2) = 10

(Deduct) -(Z/2) - 2Z = 25

or, (-Z-4Z)/2 = 25

or, -5Z = 25*2

or, Z = 50/(-5)

Therefore, Z = -10

Putting the value of Z into equation (iv) we get,

Y + 2*(-10) = -15

or, Y - 20 = -15

or, Y = 20 - 15

Therefore, Y = 5

Putting the value of Y and Z into equation (i) we get,

X + 5 + (-10) = -2

or, X + 5 - 10 = -2

or, X - 5 = -2

or, X = 5 - 2

Therefore, X = 3

So, the numbers are, X = 3; Y = 5; and Z = -10.

8 0

3 years ago

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y

Luda [366]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

$A=P(1+r)^n$ where A is number of beetles at any year n.

Plug the given values into above formula we get:

$A=100(1+0.32)^n$

$A=100(1.32)^n$

now plug n=10 years

$A=100(1.32)^{10}=100(16.0597696605)=1605.97696605$

Hence answer is approx 1606 beetles will be there after 10 years.

To find answer for 20 years plug n=20 years

$A=100(1.32)^{20}=100(257.916201549)=25791.6201549$

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

$A=100(1.32)^n$

$100000=100(1.32)^n$

$1000=(1.32)^n$

$log(1000)=n*log(1.32)$

$\frac{\log\left(1000\right)}{\log\left(1.32\right)}=n$ 24.8810001465=n

Hence answer is approx 25 years.

8 0

3 years ago

Read 2 more answers