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Ymorist [56]
3 years ago
7

6.04 x 10^-3 as an ordinary number

Mathematics
1 answer:
horrorfan [7]3 years ago
7 0
Add 3 0's because it will move the .

so the correct answer is = 0.00604
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Calcola l'area di un rombo sapendo che il perimetro è 160 cm è la sua altezza misura 38,4 cm​
AURORKA [14]

Answer:

41,6 cm

Step-by-step explanation:

Formula for Perimeter of a rhombus:  P = 2L + 2H (H is the height, L the length)

Perimeter:  160 cm = 2L + 2(38,4 cm), or

160 cm + 76.8 cm + 2L, or

83.2 cm = 2L

and so the height (altezza) is H = 83.2 cm / 2 = 41,6 cm

4 0
3 years ago
HELPP now plssss!!!!!
Bumek [7]
So go down and up then look it up on the web so you can look good in it that really helps
7 0
3 years ago
The equation y = 1/9x relates proportional quantities x and y.
ryzh [129]

Answer:

x = 36

Step-by-step explanation:

4 = 1/9 * x

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4 0
3 years ago
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⦁ Use both methods to find the distance between the points on the number line.
Vitek1552 [10]

Answer:

4 and 4

Step-by-step explanation:

Method A

1) Method A: Let 2 be the starting point and -2, the finishing one. Counting between 2 and -2, we can count a distance of 4 units. That's the simplest way, but not convenient to great numbers on the Number Line.

Method B:

There is no such thing as a negative distance, as a physical quantity. So this is the reason why we need to compute the absolute value of two numbers, which is simply what was done on Method B.

|2-(-2)|=|4|=4

As we are dealing with absolute values, the order is not relevant after all, the result remains the same. Take a look:

|-2-2|=|-4|=4

That's why the greater (2) or the lesser number (-2) can be the subtrahend (in bold within the brackets.

3 0
3 years ago
Alexander made of rectangular quilt that measured 3 1/4 feet in length by 2 3/4 feet and width to find the area multiply the len
Oksana_A [137]
\text {Area = } 3 \dfrac{1}{4}  \times 2 \dfrac{3}{4}

\text {Area = } \dfrac{13}{4}  \times \dfrac{11}{4}

\text {Area = } \dfrac{143}{16}

\text {Area = } 8 \dfrac{15}{16} \text { ft}^2
5 0
3 years ago
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