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3 years ago

Solve 4\5x-4\10=2\20

1 answer:

5 0

4/5x - 4/10 = 2/20...multiply everything by common denominator 20
16x - 8 = 2
16x = 2 + 8
16x = 10
x = 10/16 which reduces to 5/8 <=

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MIDPOINT & DISTANCE help

tamaranim1 [39]

Answer:

Step-by-step explanation:

The formula of a midpoint of (x₁, y₁) and (x₂, y₂):

$M\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

We have two points (-3, 7) and (-9, 1).

Substitute:

$x=\dfrac{-3+(-9)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{7+1}{2}=\dfrac{8}{2}=4$

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3 years ago

How do I even solve this?

mixer [17]

That’s a good question

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3 years ago

What is 2xy-3yz+5+2yz-xy

melomori [17]

To simplify this expression, we are going to combine like terms.

First, we can see that we have two terms with the variables $xy$ being multiplied together, $2xy$ and $-xy$ . Together, these add to $xy$ , so our expression is now:

$xy - 3yz + 5 + 2yz$

We also have two terms with $yz$ , $-3yz$ and $2yz$ . Together, these add to $-yz$ , so our expression is now:

$xy - yz + 5$

Our final expression is xy - yz + 5.

6 0

3 years ago

Use the discriminant to determine the number of solutions.<br> x2 + x = 12

Sati [7]

Answer:

x=4

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3 0

3 years ago

an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago