All categories

3 years ago

Is 0.05 greater than 0.083

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

no 0.05 is smaller then 0.083

Step-by-step explanation:

You might be interested in

a college graduate experts to earn a salary of $60,000 durning the first year after graduation and receive 4% raise every year a

Dima020 [189]

Hi there

The formula is
A=p (1+r)^t
A total income?
P salary 60000
R rate of increases 0.04
T time 10 years

A=60,000×(1+0.04)^(10)
A=88,814.7 round your answer
A=88815

Hope it helps

5 0

3 years ago

Find the measure of

sergij07 [2.7K]

Answer:

∠ ABC ≈ 137.9°

Step-by-step explanation:

Using the Cosine rule in Δ ABC

cos B = $\frac{a^2+c^2-b^2}{2ac}$

with a = 89, b = 144, c = 65

cos B = $\frac{89^2+65^2-144^2}{2(89)(65)}$ = $\frac{7921+4225-20736}{11570}$ = $\frac{-8590}{11570}$ , thus

B = $cos^{-1}$ ( - $\frac{8590}{11570}$ ) ≈ 137.9° ( to the nearest tenth )

4 0

4 years ago

A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running

Alborosie

Answer:

$x_1 =2 , x_2=7$

Step-by-step explanation:

Consider the revenue function given by $R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2$ . We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

$\frac{dR}{dx_1} = -10x_1-2x_2+34 =0$

$\frac{dR}{dx_2} = -16x_2-2x_1+116 =0$

From the first equation, we get, $x_2 = \frac{-10x_1+34}{2}$ .If we replace that in the second equation, we get

$-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156$

From where we get that $x_1 = \frac{156}{78}=2$ . If we replace that in the first equation, we get

$x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7$

So, the critical point is $(x_1,x_2) = (2,7)$ . We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives and check if the criteria is fulfilled in order for it to be a maximum. We get that

$\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}$

$\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16$

We have the following matrix,

$\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right]$ .

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is $(-10)\cdot (-16) - (-2)(-2) = 156>0$ and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum