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Katyanochek1 [597]
3 years ago
9

What does the product of any whole-number factor multiplied by 100 always have ? Explain

Mathematics
2 answers:
Alexxx [7]3 years ago
6 0
It has two extra zeroes because any number times one is always itself, then you would add two extra zeroes to the end. For example: 100 x 360= 36000, or 36x100=3600.
sineoko [7]3 years ago
5 0

Answer:

The product of any whole number factor multiplied by 100 will always have two zeros at the end.

Step-by-step explanation:

The product of any whole number factor multiplied by 100 will always have two zeros at the end. This is because, 100 has two zeroes and when we multiply any number by a number with zeroes, we add that number of zeroes at the number end.

Like 25\times100=2500

63\times100=6300

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A barge moves 7km/hr in still water. It travels 45km upriver and 45km downriver in a total time of 14 hr. What is the speed of t
hodyreva [135]

Answer:

Step-by-step explanation:

2(3x−5)+4x−10=4x+2(x+6)

Step 1: Simplify both sides of the equation.

2(3x−5)+4x−10=4x+2(x+6)

(2)(3x)+(2)(−5)+4x+−10=4x+(2)(x)+(2)(6)(Distribute)

6x+−10+4x+−10=4x+2x+12

(6x+4x)+(−10+−10)=(4x+2x)+(12)(Combine Like Terms)

10x+−20=6x+12

10x−20=6x+12

Step 2: Subtract 6x from both sides.

10x−20−6x=6x+12−6x

4x−20=12

Step 3: Add 20 to both sides.

4x−20+20=12+20

4x=32

Step 4: Divide both sides by 4.

4x/4 =32/4

x=8

8 0
2 years ago
GUYSSS PLSSS HELP ME WITH THIS QUESTION :))
Mrac [35]

Answer:

17

Step-by-step explanation:

1) if to solve the first inequation, tnen 3x<52; ⇔ x<52/3;

2) if to solve the seconde inequation, then 2x≥24; ⇔ x≥12.

3) according to the items 1 and 2 x∈[12;52/3);

4) the largest prime number is 17.

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2 years ago
What is the answer to <br> F(x)=x+2
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Answer:327e823

Step-by-step explanation:

7 0
3 years ago
In ΔBCA, CB = 11 cm, CG = 6 cm, AH = 9 cm. Find the perimeter of ΔBCA. Triangle BCA with inscribed circle D. Segments BF and BH,
romanna [79]

Answer:

The perimeter of Δ ABC is 40 cm ⇒ 2nd answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Circle D is inscribed in triangle ABC

- The circle touches the side AB at H , side BC at F , side CA at G

- BF and BH are tangents to circle D from point B

∴ BF = BH ⇒ tangents drawn from a point outside the circle

- CF and CG are tangents to circle D from point C

∴ CF = CG ⇒ tangents drawn from a point outside the circle

- AG and AH are tangents to circle D from point A

∴ AG = AH ⇒ tangents drawn from a point outside the circle

∵ CG = 6 cm ⇒ given

∴ CF = 6 cm

∵ CB = 11 cm ⇒ given

∵ CB = CF + FB

∴ 11 = 6 + FB ⇒ subtract 6 from both sides

∴ FB = 5 cm

∵ FB = BH

∴ BH = 5 cm

∵ AH = 9 cm ⇒ given

∵ AH = AG

∴ AG = 9 cm

∵ AB = AH + HB

∴ AB = 9 + 5 = 14 cm

∵ AC = AG + GC

∴ AC = 9 + 6 = 15 cm

∵ BC = 11 cm ⇒ given

∵ The perimeter of Δ ABC = AB + BC + CA

∴ The perimeter of Δ ABC = 14 + 11 + 15 = 40 cm

* The perimeter of Δ ABC is 40 cm

3 0
3 years ago
Cot ø undefined for ø
muminat

Answer:

cot q = x/y

Step-by-step explanation:

cot q = x/y

It's a trig function

3 0
2 years ago
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