All categories

3 years ago

If five notebooks cost $5.25 then how much would three notebooks cost

Mathematics

2 answers:

morpeh [17]3 years ago

8 0

5.25/5=1.05
1.05(3)=3.15
$3.15 for 3 notebooks

IrinaK [193]3 years ago

7 0

$3.15 I think... No promises haven't done normal math in a while.

I took 5.25 and divided it by 5 and got 1.05 and multiplied is by 3 and got $3.15

You might be interested in

Is 1/2 a rational number

Anit [1.1K]

Yes, because fractions are rational.

4 0

3 years ago

Read 2 more answers

An investor invests $500 in a certain stock. After the first six months, the value if the stock has increased at a rate of $20 p

stiv31 [10]

I cannot draw a graph so im going to explain
on the bottom u write per month and on the side and then increase the amount of money by 20$.

3 0

3 years ago

Read 2 more answers

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates

Umnica [9.8K]

Answer:

The answer is "36.14%"

Step-by-step explanation:

The complete question is given in the attached file please find it.

$\mu =41\\\\\sigma= 4\\\\P(42$

Using z-table to find the value.

$\to P(41$

This means that between 42 and 48 months, 36.14 % of scanners could be predicted will break down.

5 0

2 years ago

If a = b and b = c, then<br> a = c?

posledela

Answer:

Step-by-step explanation:

yes it a would equal c

8 0

3 years ago

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

skad [1K]

Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

Boundary points - $(4,0), (-4,0), (0,16)$

Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

6 0

3 years ago