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3 years ago

8

How can you get 2 17/20 from 2.85

1 answer:

7nadin3 [17]3 years ago

8 0

The 2 is ur whole number.
17/20...simple division = 0.85

giving u 2.85

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Here is a list of ages (years) of children in a room: 5, 8, 5, 6, 8, 3, 6 State the median.

Ierofanga [76]

Answer:

6

Step-by-step explanation:

The median of a list of numbers is the number found in the "middle" of all the numbers. What this means is that it has the same amount of numbers larger and smaller than it in the list. To find the median, we first have to organize the list from least to greatest. Doing this, we get 3, 5, 5, 6, 6, 8, 8.

Now that we have the list, we can find the "middle number" by taking off numbers the highest number then the lowest number. Doing this we get

3, 5, 5, 6, 6, 8, 8

5, 5, 6, 6, 8

5, 6, 6

6

This means our median is 6.

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3 years ago

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sp2606 [1]

Answer:

Step-by-step explanation:

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4 years ago

The difference of two number is 28. if one number is 11, find the other number

Leno4ka [110]

Answer:

39

Step-by-step explanation:

difference=28

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so:

11+28=39

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3 years ago

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In the diagram below secants PT & PU have been drawn from exterior point P such that the four arcs intercepted have the foll

fiasKO [112]

Answer:

x = 30°
RS = 30°, SR = TU = 120°, UR = 90°
∠P = 45°
∠UTS = 60°

Step-by-step explanation:

(a) If RS = x, then the sum of arcs around the circle is ...

x + 4x +4x +3x = 360°

12x = 360°

x = 30°

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(b) Based on the given ratios, the arc measures are computed from x. For example, ST = TU = 4x = 4(30°) = 120°

RS = 30°
ST = 120°
TU = 120°
UR = 90°

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(c) Angle P is half the difference of arcs TU and RS:

∠P = (TU -RS)/2 = (120° -30°)/2

∠P = 45°

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(d) Inscribed angle UTS is half the measure of the arc it intercepts. Arc RU has the measure (30° +90°) = 120°, so the measure of UTS is ...

∠UTS = 120°/2 = 60°

4 0

3 years ago

Plssss help quick i need it for sum important

Vanyuwa [196]

(a) The product of $(2x-4)\ and\ (3x^2-x+4)$ is calculated to be $=6x^3-14x^2+12x-16$

(b)The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

<u>Step-by-step explanation:</u>

<u>(a)</u>

$(2x-4)(3x^2-x+4)\\\\=(2x.3x^2)+(2x.-x)+(2x.4)+(-4.3x^2)+(-4.-x)+(-4.4)\\\\=6x^3-2x^2+8x-12x^2+4x-16\\\\=6x^3-14x^2+12x-16$

b)

$(4x-2)(3x^2-x+4)\\\\=(4x.3x^2)+(4x.-x)+(4x.4)+(-2.3x^2)+(-2.-x)+(-2.4)\\\\=12x^3-4x^2+16x-6x^2+2x-8\\\\=12x^3-10x^2+18x-8$

The product of $(2x-4) \ and\ (3x^2-x+4)\$ is not equal to the product of $(4x-2)\ and\ (3x^2-x+4)$

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3 years ago