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3 years ago

|2x−3y| if x=0.5; y=−0.7 help me out please

Mathematics

2 answers:

Inessa05 [86]3 years ago

7 0

The answer is 3.1
(2*0.5)-(3*-0.7)
1-(-2.1)
1+2.1
3.1

oee [108]3 years ago

5 0

The given expression is

$| 2x-3y | , x=0.5, y=-0.7$

Substituting the given values of x and y, we will get

$| 2(0.5)-3(-0.7) | = | 1+ 2.1 | = |3.1 |$

And absolute value of both positive and negative number is a positive number .

Therefore the final answer is

$| 3.1 | = 3.1$

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Determine the resulting polynomial: f ( x ) = 7 x + 9 f(x)=7x+9 g ( x ) = − 5 x 2 − 2 x − 8 g(x)=−5x 2 −2x−8 Find: f ( x ) ⋅ g (

S_A_V [24]

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Answer:

f(x) * g(x) = -35x^3 - 59x^2 - 74x - 72

Step-by-step explanation:

If f(x) = 7x+9 ang g(x) = -5x^2 - 2x - 8, then

f(x) * g(x) will be:

(7x+9)(-5x^2 - 2x -8)

f(x) * g(x) = -35x^3 - 59x^2 - 74x - 72

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3 years ago

9x-2y=11 5x-2y=15 what is the solution to the above system of equations ASAP

natali 33 [55]

9x - 2y = 11 ... (i)

5x - 2y = 15 ... (ii)

Subtracting equation (ii) from (i) we get;

4x + 0 = -4

4x=-4 , x = -1

Replacing x = -1 in equation (i) we get;

9(-1) - 2y = 11

-9 - 2y = 11

-2y = 20

y = 20 ÷ -2 = -10

The solution to the system of equations is (-1,-10).

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3 years ago

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Can you please answer

BabaBlast [244]

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2 years ago

Find the area of the figure

Alex17521 [72]

Answer:

To find area of parallelogram, multiply base and height

Step-by-step explanation:

Base is 6, height is 4

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3 years ago

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Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

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From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

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First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

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2 years ago

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