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faust18 [17]
3 years ago
6

What is the population of china's capital in scientific notation?

Mathematics
1 answer:
Viefleur [7K]3 years ago
6 0
<span>6.517209520 X 109 is the answer</span>
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Expand 4p(p-7a) please solve this
Inessa05 [86]
<span>Expand 4p(p-7a) 

= 4p^2 - 28ap</span>
8 0
4 years ago
Read 2 more answers
Write the equation of the line parallel to the y axis at a distance of 7 units form it to is left​
Finger [1]

Answer:

x = -7

Step-by-step explanation:

Two parallel lines are lines that never intersect each other.

The y-axis can be written as x = 0.

Then if we want a parallel line to this one, it must be something like x = b

such that b ≠ 0.

Now we want that the distance between our line and the line x = 0 is 7 units to the left

(remember that the left is the negatie side of the axis)

then we can write:

b - 0 = -7

b = -7

The parallel line is x = -7

6 0
3 years ago
Use the given information to determine if the geometric series converges or
morpeh [17]

Answe

Step-by-step explanation:

5 0
3 years ago
If x&gt;2, then x^2-x-6/x^2-4=
Viefleur [7K]
Consider the expression \frac{x^{2} -x-6}{ x^{2} -4}

To factorize the expression in the denominator we use difference of squares: x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)

To factorize x^{2} -x-6 we use the following method:

x^{2} -x-6=(x-a)(x-b)

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)

&#10; \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2}


\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}

for x>2

\frac{1}{x+2}\ \textless \  \frac{1}{2+2}= \frac{1}{4}

thus

for x>2, 

1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}


Answer: 

for x>2

\frac{x^{2} -x-6}{ x^{2} -4} =  \frac{x+3}{x+2} \ \textless \  \frac{5}{4}, (but the expression is never 0)
8 0
4 years ago
Solve the following inequality algebraically. <img src="https://tex.z-dn.net/?f=%20%7Cx%20-%201%7C%20%20%5Cleqslant%2013" id="Te
olya-2409 [2.1K]

lx-1 l ≤ 13

There are 2 solutions:

x-1 ≤ 13

and

x-1 ≥ -13

Solve each one

x-1 ≤ 13

add 1 to both sides of the equation

x-1+1≤ 13+1

x ≤ 14

x-1 ≥ -13

Add 1 to both sides:

x-1+1 ≥-13+1

x ≥ -12

-12 ≤ x ≤ 14

8 0
2 years ago
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