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Tanya [424]
3 years ago
15

What is the missing fraction for the equation? 9/10 + ?/? = 99/100

Mathematics
1 answer:
Juliette [100K]3 years ago
4 0
9/10 = 90/100, so 90/100 (which is 9/10)+ 9/100 = 99/100
You might be interested in
How to solve 3a+9 and a²-9 i have to find h.c.f
Lesechka [4]

Answer:

First equation = 3a + 9

= 3( a+3 )

Second equation = a²- 9

= a² - 3 ^ 2

= (a+3) (a-3)

Hence,

HCF = ( a+ 3)

Hope this helped

All the best !!

4 0
3 years ago
Determine whether 6 is a solution of the equation 2x - 5 = 1.
kupik [55]

Answer:

B

Step-by-step explanation:

To determine if x = 6 is a solution, substitute x = 6 into the left side of the equation and if equal to the right side then it is a solution.

2(6) - 5 = 12 - 5 = 7 ≠ 1

Then x = 6 is not a solution of the equation since a false statement results after replacing x with 6 (B)

7 0
3 years ago
What's -1.55 as a fraction? show work please help
expeople1 [14]

Answer:

hope it's help you dear

Step-by-step explanation:

The decimal part of your number seems to have the repeating digit 5 in it.

Your original number to convert is 1.550000. Let's slide the decimal point in this number to the right 1 place(s) (the same number of digits in the number 5).

If we do this, we'll get a 15.500000 (slide the decimal in the 1.550000 right 1 places, you'll get 15.500000).

So what? Well now, we have two numbers with the same repeating decimal parts, 15.500000 and 1.550000.

Now let's just work a little algebra into all of this. Let's call your original number x. And in this case, x=1.550000. The number with the decimal point slid over can be called 10x, because 10x=15.500000

What if we subtracted these two equations (that is, subtract the items on the left of the equal sign

from the stuff on the right of the equal sign)?

10x = 15.5

- x = 1.55

----------

9x = 13.95.

Now here's the important result of doing all of this: Notice how all of the repeating decimal parts have subtracted away to zero! We are left with a nice, simple 14 on the right side of the equal sign.

Now, solving 9x=14 for x by dividing both sides of it by 9, we'll get that x=14/9. And this is your answer.

How is this your answer? Well remember that above, x was originally set equal to 1.550000 via x=1.550000, and now we have that x is also equal to 14/9, so that means 1.550000=14/9..and there's 1.550000 written as a fraction!

The fraction is an improper fraction (the numerator is greater than the denominator).

While there is nothing incorrect about this, an improper fraction is typically

simplified further into a mixed number.

The whole number part of the mixed number is found by dividing the 14 by the 9.

In this case we get 1.

The fractional part of the mixed number is found by using the remainder of the division,

which in this case is 5 (14 divided by 9 is 1 remainder 5).

So your final answer is: 1.550000 can be written as the fraction

7 0
3 years ago
Read 2 more answers
If y varies directly as x and the constant variation is -2, which equation represents this relationship?
kozerog [31]

Answer:

  • y = -2x

Step-by-step explanation:

If y varies directly with x, it is:

  • y = kx

Constant is -2, it means k= -2

  • y = - 2x
4 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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