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soldier1979 [14.2K]

3 years ago

7

7/15 into a decimal rounded to the nearest hundredths place

1 answer:

hodyreva [135]3 years ago

4 0

.47 hope this helped

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URGENT HELP ME PLEASE

Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(c) $\log_2(\dfrac{64}{8})=3$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

5 0

3 years ago

Please help will give brainliest!

Ipatiy [6.2K]

Answer:

Should be 1 if im not wrong Then Mark Me brainliest

Step-by-step explanation:

a negative times a neg equals a posative

Neg x pos = neg

pos x neg = neg

pos x pos = pos

neg x neg = pos

so if there greater then over neg<, >,=

You would look at the problem for given f(x)=

x → -1 and neg times neg more or less

-1 x -1 = 1

7 0

3 years ago

HELP PLEASE! The distances from a ship to two lighthouses on the shore are 4 miles and 7 miles respectively. If the angle betwee

IrinaK [193]

Answer:

5.25 miles

Step-by-step explanation:

We are required to determine the distance between Lighthouse A and Lighthouse B in the diagram.

Using Law of Cosines

$c^2=a^2+b^2-2abCosC\\c^2=7^2+4^2-2(7)(4)Cos48^\circ\\c^2=27.5287\\c=\sqrt{27.5287} \\c=5.25\: miles$

The distance between lighthouses is 5.25 miles.

6 0

3 years ago

-11 2/3 × ( -4 1/5 )=

galina1969 [7]

The answer is 49
Solving Steps

8 0

3 years ago

PL has endpoints P(4, −6) and L(−2, 1).

masya89 [10]

I believe it'd be the one with the smiley emoji

6 0

3 years ago

Read 2 more answers