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lys-0071 [83]
3 years ago
9

Solve for n. n−10=5−4n

Mathematics
2 answers:
Alisiya [41]3 years ago
8 0
N = 3. Add 10 to both sides. Add 4n to both sides. Divide 4 from both sides. 

Julli [10]3 years ago
5 0
N - 10 = 5 - 4n
   + 10  +10

n = 15 - 4n
+4n     +4n

5n = 15

n = 3
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Answer:

True

Step-by-step explanation:

The formula for the area of a triangle is:

area=(base x height)/2

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The mass of a proton is approximately 1.67 x 10−24, and the mass of an electron is approximately 9.11 x 10−28. Find the approxim
nataly862011 [7]

Answer:

≈ 1833

Step-by-step explanation:

To find ratio of proton mass to electron mass, we have to divide.

The numbers are given in <em>scientific notation</em>.

Let a number be  a*10^b and another be  c*10^d, when we divide, we will follow the rule shown below:

\frac{a*10^b}{c*10^d}=(\frac{a}{c}*10^{b-d})

Now, we use the information given to find the ratio:

\frac{1.67*10^{-24}}{9.11*10^{-28}}\\=(\frac{1.67}{9.11}*10^{-24--28})\\=0.1833*10^4

This means we can find the number by taking 4 decimal places to the right, so that would becomes:

0.1833*10^4=1833

The approximate ratio is 1833 [mass of proton is around 1833 times heavier than mass of electron]

7 0
3 years ago
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A: 1/2 inch <br> B: 1 inch <br> C: 2 inches<br> D: 1/32 inch
mote1985 [20]
The expression of the height is

a_n=8\cdot (\frac{1}{2})^{n-1}

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An when n=5, the height is

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4 0
3 years ago
45 units and is centered at
mariarad [96]

The correct question is:

A circle has a radius of 45 units and is centered at (-2.4, -4.8).

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Answer:

Equation of the circle is;

(x + 2.4)² + (y + 4.8)² = 2304

Step-by-step explanation:

The standard equation of a circle is;

(x - a)² + (y - b)²  =  r²

where;

(a,b) is the center of the circle and r is the radius of the circle

Now, from the question, the circle is centered at (-2.4, -4.8) and the radius is 45

Thus, plugging those values into the standard form of equation of a circle, we have;

(x - (-2.4))²  +  (y - (-4.8))²  = 48²

This gives;

(x + 2.4)² + (y + 4.8)² = 2304

3 0
3 years ago
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