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german
3 years ago
5

The accepted value of the average distance between Earth and the moon is 384, 467 km. If a scientist measures that the moon is 3

84,476 km form the Earth, what is the measurement's percentage error?
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

Percentage error is 0.0024 %

Step-by-step explanation:

Initial average distance between Earth and moon = 384467 km  

Distance measure by the scientist = 384476 km

Total variation in the distance calculation = 384476 – 384467 = 9 km.

Now we can find the percentage by dividing the variation in distance from the initial measurement and then multiply with hundred.

Percentage error = ( 9 /  384467 ) × 100 = 0.0024 %

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A machine requires three hours to make a unit of Product A and five hours to
svlad2 [7]

Let A unit be a; B unit be b

a + b = 95

b = 95 - a

3a + 5b = 395

3a + 5(95 -a) = 395

3a + 475 - 5a = 395

-2a = -80

a = 40

a + b = 95

40 + b = 95

b = 55

Therefore, a = 40; b = 55.

Hope this helps

8 0
3 years ago
Please look at picture
Alex_Xolod [135]
6x - 7 = 4x + 7
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x = 7

The answer is 7
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3 years ago
Graph the relation and its inverse. Use open circles to graph the points of the inverse. x –3 4 6 9 y 5 6 –9 –10
Tema [17]

Answer:

See attached picture.

Step-by-step explanation:

Graph the function as (x,y) points.

(-3,5)

(4,6)

(6,-9)

(9,-10)

These are graphed in black on the picture.

To graph the inverse, switch the points from (x,y) to (y,x).

(5,-3)

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(-9,6)

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These are graphed in red on the picture.

4 0
3 years ago
Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
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andrey2020 [161]
I will assume that this question is about scientific notation (also assuming that the first number is 3.4*10^2), where you can arrange the numbers by the exponent of the 10, so from least to greatest:
8.11*10^{-3}, 3.4*10^{2}, 435, and 1.2*10^{7}
6 0
3 years ago
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