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3 years ago

11

5 less than two times a number is equal to 4 less than the same number.

2 answers:

kati45 [8]3 years ago

3 0

Answer:

x=1

Step-by-step explanation:

2x-5=x-4

x=1

Montano1993 [528]3 years ago

3 0

Answer:

The number is 1

Step-by-step explanation:

Let the number be x

2(x) - 5 = x - 4

2x - 5 = x - 4

2x - x = -4 + 5

x = 1

Therefore the number is 1

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What is the multiplicative rate of change for the exponential function f(x) = 2()–x?

Sauron [17]

I think its 2/5 = .4 <span>because of the negative in the exponent, it is 0.4</span> Hope I helped :)

4 0

3 years ago

Read 2 more answers

three adults and four children must pay $122 for tickets. Two adults and three children must pay $87. what are the price of each

Natasha2012 [34]

Let $a$ be the price of an adult ticket, and $c$ the price of a child ticket.

The sentence "Three adults and four children must pay $122 for tickets" translates to $3a+4c = 122$

The sentence "Two adults and three children must pay $87" translates to $2a+3c = 87$

Which leads to the linear system

$\begin{cases} 3a+4c = 122\\2a+3c = 87 \end{cases}$

You can solve this system as you prefer, for example you can multiply the first equation by 2 and the second by 3 to get

$\begin{cases} 6a+8c = 244\\6a+9c = 261\end{cases}$

Now subtract the first from the second:

$(6a+9c) - (6a+8c) = 261-244 \iff c = 17$

Now plug this value for c in any of the equations, for example the first:

$3a+4c = 122 \iff 3a+4\cdot 17 = 122 \iff 3a+ 68=122 \iff 3a = 54 \iff a = 18$

5 0

3 years ago

plz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help i need helpplz help

Volgvan

Answer:

the answer should be 9

Step-by-step explanation:

$(256 * 3^{-5} * 6^{0} )^{-2} * (\frac{3^{-2}}{2^{3}} )^{4} * 2^{28}$

$(256*\frac{1}{3^{5}} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}$

$(256*\frac{1}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}$

$(\frac{256}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}$

$(\frac{256}{243})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}$

$(\frac{243}{256})^{2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}$

$\frac{59049}{65536}*(\frac{1}{2^{3}*3^{2}})^{4}*2^{28}$

$\frac{59049}{65536}*(\frac{1}{8*9})^{4}*2^{28}$

$\frac{59049}{65536}*(\frac{1}{72})^{4}*2^{28}$

$\frac{59049}{65536}*\frac{1}{26873856}*2^{28}$

$\frac{9}{65536}*\frac{1}{4096}*2^{28}$

$\frac{9}{65536*4096}*2^{28}$

$\frac{9}{268435456}*268435456$

$9$

6 0

3 years ago

Read 2 more answers

I’ll give points + brainalist for the correct answer (:

ludmilkaskok [199]

JENNA

C: is my answer cuz equivakient are the same in fraction..

Step-by-step explanation:

Is this correct let me know!!

8 0

3 years ago

What is the missing product<br> 4, 6, 9,<br> 14, ?, 49

ser-zykov [4K]

I think it's 35, because 49-14=35

3 0

3 years ago