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aleksandrvk [35]
3 years ago
12

What is (4×3)÷2+1+2×6

Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0

Answer:

19

Step-by-step explanation:

(4*3)/2+1+2*6

=12/2+1+2*6

=6+1+2*6

=6+1+12

=19

Olenka [21]3 years ago
3 0

Answer:

(4×3)÷2+1+2×6

=12÷2+1+2×6

=6+1+2×6

=6+1+12

=19

is the answer

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A fluid moves through a tube of length 1 meter and radius r=0.006±0.00025 meters under a pressure p=4⋅105±2000 pascals, at a rat
iris [78.8K]

Answer:

Maximum error for viscosity is 17.14%

Step-by-step explanation:

We know that everything is changing with respect to the time, "r" is changing with respect to the time, and also "p" just "v" will not change with the time according to the information given, so we can find the implicit derivative with respect to the time, and since

n = (\frac{\pi}{8}) (\frac{pr^4}{v})\\

The implicit derivative with respect to the time would be

\frac{dn}{dt}  = \frac{\pi}{8} ( \frac{r^4}{v} \frac{dp}{dt}  + \frac{4pr^3}{v}\frac{dr}{dt} )

If we multiply everything by    dt   we get  

dn  = \frac{\pi}{8} ( \frac{r^4}{v} dp  + \frac{4pr^3}{v}  dr})

Remember that  the error is given by   \frac{dn}{n}   therefore doing some algebra we get that

\frac{dn}{n}  =    4 \frac{dr}{r}  +  \frac{dp}{p}

Since,    r = 0.006   ,   dr = 0.00025 ,  p = 4*105   ,   dp = 2000  we get that

\frac{dn}{n} = 0.1714

Which means that  the maximum error for viscosity is 17.14%.  

5 0
2 years ago
For f(x)=4x+1 and g(x)=x^2-5, find (g/f)(x)
ddd [48]

Answer:

(g/f)(x) =  \frac{g(x)}{f(x)}  =  \frac{ {x}^{2} - 5 }{4x + 1}

I hope I helped you^_^

3 0
2 years ago
Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Match each set of vertices of quadrilateral EFGH with t
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Each transformation will give these following coordinates

Translation 7 units right gives E(4, 4), F(8, 3), G(10, 6), and H(8, 6) as shown in brown in the diagram below

Reflection on the y-axis gives E(3, 4), F(-1, 3), G(-3, 6) and H(-1, 6) as shown in green in the diagram below

Reflection on the x-axis gives E(-3, -4), F(1, -3), G(3, -6) and H(1, -6) as shown in red in the diagram below

Translation 5 units down gives E(-3, -1), F(1, -2, G(3, 1) and H(1, 1) as shown in purple in the diagram below 

8 0
3 years ago
Find the distance between the two points (-10,3) and (0,27)​
crimeas [40]

Answer:

26 units

Step-by-step explanation:

\boxed{distance \: between \: 2 \: points =  \sqrt{ {(x1 - x2)}^{2} +  {(y1 - y2)}^{2}  } }

Using the formula above,

distance between (-10, 3) and (0, 27)

=  \sqrt{ {( - 10 - 0)}^{2}  +  {(3 - 27)}^{2} }

=  \sqrt{ {10}^{2}  + ( - 24) {}^{2} }

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5 0
2 years ago
Which expression is equivalent to (x^4/3 x^2/3)^1/3?
KATRIN_1 [288]

Answer: x^{\frac{2}{3} }

Step-by-step explanation:

We have several properties of exponents in use here. The two that are used are:

(x^{a})(x^{b}) = x^{a + b} <em>(Exponents with the same base that are being multiplied together can have the exponents added)</em>

(x^{a})^{b} = x^{(a)(b)} <em>(A base raised to a power, and then raised to another power means that you can multiply the exponents to get the same result as doing inside operations and then outside operations)</em>

<em />

Let's apply it!

First, let's simplify what's inside the parenthesis.

x^{\frac{4}{3} } x^{\frac{2}{3} } <em>(Remember, they have the same base of "x", so we can add the exponents)</em>

x^{\frac{4}{3} + \frac{2}{3} } = x^{\frac{6}{3} } = x^{2}

Now we have (x^{2})^{\frac{1}{3} }. Let's use the second rule.

(x^{2})^{\frac{1}{3} } = x^{\frac{2}{3} }

Hope this helps! :^)

7 0
3 years ago
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