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evablogger [386]

3 years ago

12

The temperature in a chemical reactor was measured every half hour under the same conditions. The results were 78.1°C, 79.2°C, 7

8,9°C, 80.2°C, 78.3°C, 78.8°C. 79.4°C.
the mean ?

1 answer:

gladu [14]3 years ago

3 0

Answer:

72.2

Step-by-step explanation:

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Enter an equation that can be used to solve this problem, where x represents the number of hours that the first plane had been t

Tom [10]

X=number of hours that the first plane had been traveling.
(x-2)=number of hours that the second plane had been traveling.

we can suggest this equation:

500x+550(x-2)=3100

Now, we can solve this equation:
500x+550(x-2)=3100
500x+550x-1100=3100
1050x=3100+1100
1050x=4200
x=4200/1050
x=4

Answer:
Equation: 500x+550(x-2)=3100
The amount hours, after the first plane left , when the planes will be 3100 miles would be 4 hours.

3 0

3 years ago

Rachel buys 30 nuts and bolts to finish a project. If each but costs 10 cents, each bolt costs 20 cents and the total purchase p

Natali5045456 [20]

Answer:

12 bolt 18 nuts

Step-by-step explanation: (Elimination method)

N=nuts

B=bolts

B + N = 30

0.20B + 0.10N = 4.20

20B+10N=420 (I multiplied by 10 to eliminate decimals

-------------------------------------------------------

I will multiply the first by (-10)

-10B -10N =-300

20N +10N=420

10B+0 = 120

B=12 (12 BOLT)

30-12 =18 (8 NUTS)

12*0.20=2.40

18*0.10=1.80

2.40+ 1.80 = 4.20

3 0

2 years ago

Sam is 11 years old. Ross is twice as old as Sam. When Ross is 50 years old, how old will Sam be?

svlad2 [7]

Answer:

61

Step-by-step explanation:

When Sam is 11 Ross is 22 which is +11 of Sam's age

So when Ross is 50 Sam will be 50+11= 61

3 0

3 years ago

the terminal side of 0 passes through the point (8,-7) what is the exact value of cos 0 in simplified form

Genrish500 [490]

Answer: $\cos(\theta) = \frac{8\sqrt{113}}{113}\\\\$

=================================================

Work Shown:

x^2+y^2 = r^2

(8)^2+(-7)^2 = r^2

113 = r^2

r = sqrt(113)

The distance from (0,0) to (8,-7) is exactly sqrt(113) units.

This is the exact length of the hypotenuse of the right triangle.

Next, we do the following steps:

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(\theta) = \frac{x}{r}\\\\\cos(\theta) = \frac{8}{\sqrt{113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113}*\sqrt{113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113*113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113^2}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{113}\\\\$

Side note: cosine is positive in quadrant Q4.

6 0

3 years ago

Compute the matrix of partial derivatives of the following functions.

s344n2d4d5 [400]

For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$

(c)

$Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}$

(d)

$Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}$

5 0

3 years ago