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3 years ago

How many times does 6 go into 96 by steps

Mathematics

2 answers:

guapka [62]3 years ago

8 0

96 ÷ 6 = 16

It might be easier to do if you have a calculator at hand, but if not, here is a picture of how it's done:

dangina [55]3 years ago

4 0

6,12,18,24,30,36,42,48,54,60,66,72,78,84,90, 96
So 6 goes into 96 16 times
96÷6=16

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What is the value of x?

Serjik [45]

Answer:

x = 32

Step-by-step explanation:

Each angle of the equilateral triangle is 60°.

(2x -4)° = 60°

x -2 = 30 . . . . . . divide by 2°

x = 32

5y° = 60°

y = 12 . . . . . . . . . divide by 5°

7 0

3 years ago

Can anybody help me please? i dont get it :(

sweet [91]

Answer:

ay bro dont ask me

the answer above me is already correct

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Step-by-step explanation:

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2 years ago

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

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

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2 years ago

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