All categories

2 years ago

? ÷ 100,000 = 4.3976 ? ÷ 100,000 = 439.76 ? ÷ 100,000 = 4,397.6 ? ÷ 100,000 = 439,760

Mathematics

1 answer:

Olin [163]2 years ago

4 0

Answer:

1 = 439,760

2= 43,976,000

3= 439,760,000

4= 43,976,000,000

Step-by-step explanation:

You can easily get the answers by multiplying each by 100,000

You might be interested in

Imagine a pond. In it sits one lilypad, which reproduces once a day. Each of its offspring also reproduces once a day, doubling

sergeinik [125]

Answer: On the 29th day

Step-by-step explanation:

According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.

On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.

The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are $2^{30}$ lilypads on the 30th, and $2^{29}$ lilypads on the 29th. This can be deduced multiplying succesively by 2.

4 0

3 years ago

A need help Dilations-He Said, She Said (12/10)

Elodia [21]

Answer:

We conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

Step-by-step explanation:

Checking the location of the vertices of the original triangle ΔDEF

D(-2, 4)
E(2, 4)
F(0, 2)

Checking the location of the dilated vertices of the image triangle ΔD'E'F'

D'(-5, 10)
E'(5, 10)
F'(0, 5)

Notice that if we multiply the vertices of the original triangle ΔDEF by 5/2, we get the correct image vertices of image triangleΔD'E'F'

i.e.

D(-2, 4) → D'(5/2 (-2), 5/2 (4)) → D'(-5, 10)

E(2, 4) → E'(5/2 (2), 5/2(4)) → D'(5, 10)

F(0, 2) → E'(5/2 (0), 5/2(2)) → F'(0, 5)

Thus, we conclude that the original triangle ΔDEF is dilated by a scale factor of 5/2 to get the image triangle ΔD'E'F'.

Therefore, 'Cárl' is correct. Hence, option 'a' is correct.

6 0

2 years ago

The point A(-8, 6) is translated using T: (x,y) → (x + 5. y - 4). What is the distance from A to A'?

Vsevolod [243]

Translated means the points are moving across the plane without rotating or changing shape. In this case, the x-coordinate would be moving up 5 (x + 5) and the y-coordinate would be moving to the left 4 (y - 4).

A is (-8, 6). A' is the result of the translation from this point. The results of the solution above in A is the point (-3, 2) = A'.

Now you must find the distance between these two coordinates. To find the distance you must use the distance formula: √(x2 - x1)^2 + (y2 - y1)^2. Since you now have two points, A and A', plug these into the distance formula.

√(-3 - (-8))^2 + (2 - 6)^2
√5^2 + (-4)^2
√25 + 16
√41

The distance from A to A' is √41.

8 0

3 years ago

Read 2 more answers

Is 5.564 bigger than 5.654

Charra [1.4K]

Answer:

Step-by-step explanation:

5.564 - 5.654= -0.09

----------------------

8 0

2 years ago

Read 2 more answers

F(x)=20-x^2+8x what are the minimum and maximum points and the axis of symmetry?

-Dominant- [34]

Vertex aka max or min point is found by -b/2a in form
f(x)=ax^2+bx+c
f(x)=-1x^2+8x+20
vertex x value is -8/(2)(-1)=-8/-2=4
input back to find y value
f(4)=-(4^2)+8*4+20
f(4)=-16+32+20
f(4)=36
max (since the graph opens down) is (4,36)
axis of symmetry is the x coordinate

max is (4,36)
axis of symmetrry is x=4

5 0

2 years ago