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IrinaK [193]
4 years ago
9

Which would the phrase twice the difference of a number and five look like as a variable expression

Mathematics
1 answer:
joja [24]4 years ago
4 0

We have to represent the phrase "twice the difference of a number and five" as variable expression.

A variable expression is a mathematical phrase that can contain ordinary numbers, variables (like x or y) with operations as addition, subtraction ,multiplication and division.

Let the number be 'x'.

Twice of a number = 2 \times x = 2x

The difference of '2x' and 5 = 2x - 5

Therefore, the variable expression representing the given phrase is "2x-5".

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dybincka [34]

Answer:

y=2×+1 im a bit rusty hope this helps in some way

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3 years ago
In triangleABC, C is the right angle. If tanA = 8/6, find cosB
klemol [59]

<u>Short answer</u> = C) 8/10

<u>Working :</u>

Tan A = 8/6 = Opposite / AdjacentOpposite = 8 units

Adjacent = 6 units

Hypotenuse = \sqrt{8^{2} + 6^{2} }

                     = \sqrt{64 + 36 }

                     = \sqrt{100}

                     = 10 units

Cos B  = Adjacent/ Hypotenuse

        ∴  cos B= 8/10

4 0
4 years ago
Are the vectors equal?
Lesechka [4]

Answer: It depends

Step-by-step explanation:

They are equal when they have the same length and point in the same direction. Therefore, it depends. Hope this helps! I'm sorry if I'm wrong.

-Bella :)

8 0
3 years ago
A market trader asks 500 for some cloth. a woman offers 1200, after bargaining they agree a price half way betweeb the two start
Arturiano [62]

Answer:

850

Step-by-step explanation:

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6 0
3 years ago
Suppose the radius of the sphere is increasing at a constant rate of 0.3 centimeters per second. At the moment when the radius i
elixir [45]
<h2>At the moment when the radius is 24 centimeters, the volume is increasing at a rate of 2171.47 cm³/min.</h2>

Step-by-step explanation:

We have equation for volume of a sphere

             V=\frac{4}{3}\pi r^3

where r is the radius

Differentiating with respect to time,

            \frac{dV}{dt}=\frac{d}{dt}\left (\frac{4}{3}\pi r^3 \right )\\\\\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2\times \frac{dr}{dt}\\\\\frac{dV}{dt}=4\pi r^2\times \frac{dr}{dt}

Given that

           Radius, r = 24 cm

           \frac{dr}{dt}=0.3cm/s

Substituting

           \frac{dV}{dt}=4\pi r^2\times \frac{dr}{dt}\\\\\frac{dV}{dt}=4\pi \times 24^2\times 0.3\\\\\frac{dV}{dt}=2171.47cm^3/min

At the moment when the radius is 24 centimeters, the volume is increasing at a rate of 2171.47 cm³/min.

4 0
3 years ago
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