What is the length of the diagonal from vertex J to vertex L in the quadrilateral below? A. √113 B. 11 units C. √103 D. 3 √33

Mathematics

1 answer:

sdas [7]3 years ago

6 0

Answer:

the answer is B.

Step-by-step explanation:

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Step by step workout of ; log x^3+log 5x=5 log 2-log 2/5

Charra [1.4K]

Answer:

x = 2

Step-by-step explanation:

$log \: {x}^{3} + log \: 5x = 5log \: 2 - log \: \frac{2}{5} \\ \\ \therefore \: log( {x}^{3} \times 5x) = log {2}^{5} - log \: \frac{2}{5} \\ \\ \therefore \: log( 5{x}^{4}) = log 32 - log \: \frac{2}{5} \\ \\ \therefore \: log( 5{x}^{4}) = log \bigg(32 \times \frac{5}{2} \bigg) \\ \\ \therefore \: log( 5{x}^{4}) = log \bigg(16 \times 5) \\ \\ \therefore \: 5{x}^{4}= 16 \times 5 \\ \\ \therefore \: {x}^{4}= {2}^{4} \\ \\ \huge \purple{ \boxed{\therefore \: {x} = 2}}$

4 0

3 years ago

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. In y

BabaBlast [244]

F(x)=2ˣ
f(x)
(x,y)
(0,1)
(1,2)
(2,4)
(3,8)
(4,16)
(5,32)
(6,64)
(7,128)
(8,256)
(9,512)
(10,1024)

h(x)
(x,y)
(0,8)
(1,10)
(2,18)
(3,38)
(4,76)
(5,138)
(6,230)
(7,358)
(8,528)
(9,736)
(10,1018)

so somwehre between x=9 and x=10

hmm
f(9.5)=724.007
h(9.5)=874.875

go to 9.75
f(9.75)=944.609
h(9.75)=861.078

up

9.9
f(9.9)=955.426
h(9.9)=988.199
up

9.95
f(9.95)=989.119
h(9.95)=1003.02
up

9.99
f(9.99)=1016.93
h(9.99)=1014.99

a bit down
9.98
f(9.98)=1009.9
h(9.98)=1011.99

a bit up
9.985
f(9.985)=1013.41
h(9.985)=1013.49
pretty close, ya

so about at 9.985=x
if we graphed, we get that x=9.98512070914982..., so we are pretty close

8 0

3 years ago

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Anton [14]

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Step-by-step explanation:

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