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2 years ago

Please help with this question!!

Mathematics

1 answer:

mars1129 [50]2 years ago

7 0

Answer:

second one

Step-by-step explanation:

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V-9/10v+6=11(hlp)this problem is confusing.

Valentin [98]

$\frac{v-9}{10v+6}=11 \\\\ 11(10v+6)=v-9 \\\\ 110v+66=v-9 \\\\ 110v-v=-9-66 \\\\ 109v= -75 \\\\ \boxed{v=-\frac{75}{109}}$

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3 years ago

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Which statement is true?

g100num [7]

A is the answer to the equation

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2 years ago

Consider the arithmetic sequence:

prisoha [69]

Answer:

Answer: D

Step-by-step explanation:

Arithmetic Sequences

The arithmetic sequences are those where any term n is obtained by adding or subtracting a fixed number to the previous term. That number is called the common difference.

The equation to calculate the nth term of an arithmetic sequence is:

$a_n=a_1+(n-1)r$

Where

an = nth term

a1 = first term

r = common difference

n = number of the term

We are given the arithmetic sequence:

10, 12, 14, 16, ...

Where a1=10 and

r = 12 - 10 = 2

Thus the general term is:

$a_n=10+(n-1)2$

Operating:

$a_n=10+2n-2$

$a_n=8+2n$

Answer: D

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3 years ago

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The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and

NemiM [27]

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

$60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120$

Let's find h with the quadratic formula:

$\fbox {Quadratic formula:} \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}$

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation

$\fbox {Pythagorean equation:}$ The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle).