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5 and 3/8 + 2 and 2/8

zmey [24]

7and 5/8
5+2 is 7
3/8+2/8 is 5/8
7+5/8 is
7 and 5/8

4 0

3 years ago

Read 2 more answers

The answer is not A can someone plz help me I’m not understanding it thank you it would be appreciated

jeka94

nswer:

209.8

Step-by-step explanation:

the bottom is = 11 * 5

two of the sides are 11 * 9.3 / 2 because they are triangles, and then times 2 because there are two of them

the other two sides are 10.5 * 5 / 2 , and the * 2 because there are two of them again.

(11 * 5) + (11 * 9.3 * 2 / 2) + (10.5 * 5 * 2 / 2) =209.8

6 0

3 years ago

#7-9 please help with an explanation,, will mark brainliest

Mandarinka [93]

Answers:

Step-by-step explanation:

To find out which is greater, we must eliminate the radical symbol.

#7

$\sqrt{88}$ = 9.38083151965

9.4 < 12

#8

$-\sqrt{18}$ = -4.24264068712

-4.2 > -6

#9

$\sqrt{220}$ = 14.8323969742

14.5 < 14.8

I'm always happy to help :)

8 0

3 years ago

What is 1 and 1/2 worth

koban [17]

1 is a whole and 1/2 is just a fraction

4 0

3 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

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

4 0

2 years ago