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3 years ago

Subtract equation two 3 digit numbers with different hundreds digits and a difference of less than 100

Mathematics

1 answer:

klio [65]3 years ago

6 0

Answer:

690 - 595

Step-by-step explanation:

As we know that representation of a 3 digit number = xyz

where x $\neq$ 0

z is at ones place

y is at tens place

x is at hundred place

As, we have to subtract two 3 digits numbers with different hundred digits

So, Let

The hundred digit of first number be 6

The hundred digit of second number be 5

∴

Assume first number with hundred digit 6

Let the number be - 690

Now,

Assume second number with hundred digit 5

Let the number be - 595

Now,

We will check that the difference between both the numbers are less than 100 or not

Now,

690 - 595 = 95 < 100

It satisfies the condition

So, the equation becomes 690 - 595

Read more about expressions at:

brainly.com/question/723406

#SPJ1

6 0

2 years ago

The midpoint of segment AB is (4, 2). The coordinates of point A are (2, 7). Find the coordinates of point B.

erastovalidia [21]

Answer:

<h3>The option B) is correct</h3><h3>Therefore the coordinate of B $(x_2,y_2)$ is (6,-3)</h3>

Step-by-step explanation:

Given that the midpoint of segment AB is (4, 2). The coordinates of point A is (2, 7).

<h3>To Find the coordinates of point B:</h3>

Let the coordinate of A be $(x_1,y_1)$ is (2,7) respectively
Let the coordinate of B be $(x_2,y_2)$
And Let M(x,y) be the mid point of line segment AB is (4,2) respectively
The mid-point formula is

Now substitute the coordinates int he above formula we get
$(4,2)=(\frac{2+x_2}{2},\frac{7+y_2}{2})$
Now equating we get

$4=\frac{2+x_2}{2}$ $2=\frac{7+y_2}{2}$

Multiply by 2 we get Multiply by 2 we get

$4(2)=2+x_2$ $2(2)=7+y_2$

$8=2+x_2$ $4=7+y_2$

Subtracting 2 on both

the sides Subtracting 7 on both the sides

$8-2=2+x_2-2$ $4-7=7+y_2-7$

$6=x_2$ $-3=y_2$

Rewritting the above equation Rewritting the equation

$x_2=6$ $y_2=-3$

<h3>Therefore the coordinate of B $(x_2,y_2)$ is (6,-3)</h3><h3>Therefore the option B) is correct.</h3>

7 0

3 years ago

Find the product. (4/5)(9/5)(-1/2)

MatroZZZ [7]

I believe it’s -0.72

5 0

3 years ago

Read 2 more answers

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

3 years ago