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3 years ago

11

What is 0.012345679 as a fraction in simplest form

1 answer:

gavmur [86]3 years ago

7 0

12345679/100000000 I think

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Harry had $32. He spent all the money on buying 3 notebooks for $x each and 4 packs of index cards for $y each. If Harry had bou

sattari [20]

Answer:

a. The student's conclusion is incorrect because the solution to the system of equations 3x + 4y = 32 and 5x + 5y = 50 is (8, 2).

Step-by-step explanation:

The second equation needs to be ...

5x +5y = 50

because the wording "short of $18" means that the actual cost of that purchase would be $18 more than the $32 that Harry had. 18+32 = 50.

The only answer choice that shows this as the second equation is choice A.

__

In fact, the solution to the system of equations is (8, 2), as can be seen in the graph below.

4 0

4 years ago

Round 0.0695712 to 2 significant figures

ddd [48]

0.070 is rounded to 2 significant figures

6 0

3 years ago

If 9x+10y=2 and -8x-6y=-10 are true equations, what would be the value of x + 4y?

qwelly [4]

Use elimination method to get -8

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3 years ago

What is the probability of rolling a 3 on two consecutive rolls of a fair die

n200080 [17]

So on one die has six numbers, the probability of rolling a 3 on one die is 1/6. So for two die, it would be 1/6 times 1/6= 1/36...

5 0

3 years ago

The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

4 0

3 years ago