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2 years ago

Theo has one piece of train track that is 74 millimeters long and a second piece of train track that is 97 millimeters long.

Mathematics

2 answers:

ziro4ka [17]2 years ago

8 0

Answer:

171 m i hop this helps

Step-by-step explanation:

Lana71 [14]2 years ago

7 0

Theo has 171 m of track.

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malfutka [58]

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3 years ago

Read 2 more answers

Please help FAST I need the answer along with you explaining the steps so I can get it thanks! I’ll give Brainlist also!

kupik [55]

Answer:

The equation of required line is: $\mathbf{5x-8y+34=0}$

Step-by-step explanation:

We need to write equation in the form Ax+By+C=0 of the line parallel to $5x-8y+12=0$ and through the point (-2,3)

First we need to find slope and y-intercept of the required line.

Using equation of line $5x-8y+12=0$ to find slope.

Since the given line and required lines are parallel there slope is same.

Writing equation in slope intercept form: $y=mx+b$ where m is slope.

$5x-8y+12=0 \\-8y=-5x-12\\\frac{-8y}{-8y}=\frac{-5x}{-8}-\frac{12}{-8}\\y=\frac{5}{8}x+\frac{3}{4}$

So, the slope m = 5/8

The slope of required line is $m=\frac{5}{8}$

Now finding y-intercept using slope $m=\frac{5}{8}$ and point(-2,3)

$y=mx+b\\3=\frac{5}{8}(-2)+b\\3=\frac{-5}{4}+b\\b=3+\frac{5}{4}\\b=\frac{3*4+5}{4}\\b=\frac{12+5}{4}\\b=\frac{17}{4}$

So, the equation of required line having slope $m=\frac{5}{8}$ and y-intercept $b=\frac{17}{4}$

$y=mx+b\\y=\frac{5}{8}x+\frac{17}{4}\\y=\frac{5x+17*2}{8}\\y= \frac{5x+34}{8} \\8y=5x+34\\5x-8y+34=0$

So, The equation of required line is: $\mathbf{5x-8y+34=0}$

3 0

1 year 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