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3 years ago

I need help on my homework

Mathematics

1 answer:

emmainna [20.7K]3 years ago

8 0

Same but what subject do you need help

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Una araña comió 25%, percent más bichos este mes que el mes pasado. La araña comió 8 bichos el mes pasado

lesantik [10]

For this case we have that the complete question is:

A spider ate 25% more bugs this month than last month. The spider ate 8 bugs last month. How many bugs did the spider eat this month?

So, we propose a rule of three:

8 bugs -----------> 100%

x -----------------------> 25%

Where the variable "x" represents the number of bugs equivalent to 25%.

$x = \frac {25 * 8} {100}\\x = 2$

Thus, 2 bugs represent 25%

If this month the spider ate 25% more, then it ate:

$8 + 2 = 10$

Answer:

This month, the spider ate 10 bugs

7 0

4 years ago

Read 2 more answers

Compare multiplying a number by 3 and by 1 third. *

Snowcat [4.5K]

Answer:

Multyplying by 1 third is the same as dividing by 3, so it's the opposite as multiplying by 3.

6 0

3 years ago

blake is signing a lease for a studio apartment that rents for $425/month. how much will blake pay to move in if he must pay the

Maslowich

So he would have to pay the first and last months rent so you would take 425*2=850. And then 50% of 425 would be 212.50. So all together he would be paying $1062.50 to move in. I hope this helps :)

6 0

3 years ago

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[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation

irina1246 [14]

Answer:

$y=-\frac{158}{579}$

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

$A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]$

And the vector B is formed with the solution of each equation of the system: $b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]$

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called $A_{2}$ .

$A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]$

The value of y using Cramer's rule is:

$y=\frac{det(A_{2}) }{det(A)}$

Find the value of the determinant of each matrix, and divide:

$y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}$