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3 years ago

A right cylinder has a diameter of 30 mm and a height of 30 mm, as shown.

1 answer:

3 0

"As shown" ? You have a picture to work with but you're not sharing it ?

Fortunately, the question includes all the information that's needed to answer it,
so I don't need to see the picture.

The volume of a cylinder is (pi) x (radius)² x (height) .

Radius = 1/2 of the diameter = 15 mm

Volume = (pi) x (15 mm)² x (30 mm) = 6,750 pi mm³ = 21,205.75 mm³ .

That's a correct answer, but the number is slightly awkward and inconvenient. Let's
change it to a bigger unit, so the number will be smaller and easier to carry around.

Remember that 10 mm = 1 cm

So (10 mm)³ = 1,000 mm³ = 1 cm³

and 21,205.75 mm³ = 21.20575 cm³

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If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

3 years ago

Austin keeps a right conical basin for the birds in his garden as represented in the diagram. The basin is 40centimeters deeps a

Gwar [14]

Answer:

51.11 cm

Step-by-step explanation:

hello i will help you

7 0

3 years ago

$14.65 plus what equals $65.00?

ziro4ka [17]

79.65

I used a calculator

5 0

2 years ago

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Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$

3 0

3 years ago

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5x + 7x + 2x + 8 -2 please helppp

Murljashka [212]

Answer:

= 14X + 6

Step-by-step explanation:

5 0

2 years ago

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