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2 years ago

What is the differences between normal and perpendicular?

1 answer:

6 0

Perpendicular means when a segment goes from the middle of another segment and is called the perpendicular foot and it goes like this
|
_____|_____ =>these two are perpendicular

Sorry for my bad demonstration

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Si 4x= 5/6y entonces 5y=

Hunter-Best [27]

Answer:

$5y=24x$

Step-by-step explanation:

<em>The question in English is</em>

If 4x= 5/6y then 5y=

we have

$4x=\frac{5}{6}y$

Solve for 5y

Multiply both sides by 6

$6*4x=(6)*\frac{5}{6}y$

Simplify

$24x=5y$

Rewrite

$5y=24x$

8 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

Sonia used $12.50 to buy a new journal. She has $34.25 left in her savings account. How much money did Sonia have before she bou

natali 33 [55]

X-12.50=34.25

Add 12.50 to both sides
X=46.75

3 0

3 years ago

Read 2 more answers

amid has gained weight he now weighs 88kg which is 10% higher than his normal weight. what is his normal weight?

dybincka [34]

Answer:

88*0.9= 79.2 (since it's 10% higher, multiply by .9 to find out what 90% of 88 is

you also could've done 88*0.1=8.8 and then subtract 8.8 from 88 and get 79.2

either way his normal weight is 79.2kg

5 0

2 years ago

What is the first step when solving this equation?

notka56 [123]

Answer:

Distribute

When you have an equation like this, distribute first and then add or subtract like terms to solve for x.

Hope this helps!!

6 0

3 years ago

Read 2 more answers