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3 years ago

2^5 over 4 = 2^5 over 2a = 2^b = c What are a b and c

Mathematics

1 answer:

likoan [24]3 years ago

6 0

Answer:

a = 2

b = 3

c = 8

Step-by-step explanation:

2^5/4=2^5/2a=2^b=c

2^5/4, 2^5 = 32, 32/4 = 8

so c, being what all of this is equal to, is 8

2^5/2a, 2*2=4

so a = 2

2^b = 8, ∛8 = 3

so b = 3

Hope my explanation makes sense :)

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4/3 t = 9 i need help with this question

Sonja [21]

Answer:

t=27/4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The function f(x) = 6x + 8 is transformed to function g through a horizontal stretch by a factor of 5. What is the equation of f

elixir [45]

Answer:

$g(x) = 30\cdot x +40$ ; $a \approx 1.898$ , $k = 40$ .

Step-by-step explanation:

The resultant function is obtained by multiplying $f(x)$ by a real number $k$ . That is:

$g(x) = k \cdot f (x)$

If $k = 5$ and $f(x) = 6\cdot x + 8$ , then $g(x)$ is:

$g(x) = 5\cdot (6\cdot x + 8)$

$g(x) = 30\cdot x +40$

Given that presence of the expression $g(x) = 6^{a}\cdot x + k$ , then:

$6^{a} = 30$ and $k = 40$

The value of a is obtained by applying the definition of logarithms:

$a = \log_{6}30$

$a \approx 1.898$

Finally, the value of k is found by direct comparison:

$k = 40$

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3 years ago

If f(x)=x-3, which inequality can be used to find the domain of f(x)?

Scrat [10]

Given that,

$f(x) = \sqrt{x-3}$

To find,

The domain of f(x).

Solution,

We have, $f(x) = \sqrt{x-3}$

To find domain,

The term under the square root must be positive or null as follows :

$x-3\ge 0$

Adding 3 both sides

$x-3+3\ge 0+3\\\\x\ge 3$

So, the domain is x-3>0 .

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2 years ago

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Ksenya-84 [330]

Answer:

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Step-by-step explanation:

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2 years ago

Solve 2(b − 3) ≥ 3(3 − b) and describe the graph of the solution. Show all your work.

solong [7]

Answer:

$\large\boxed{b\geq3\to b\in[3,\ \infty)}$

Step-by-step explanation:

$2(b-3)\geq3(3-b)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(2)(b)+(2)(-3)\geq(3)(3)+(3)(-b)\\\\2b-6\geq9-3b\qquad\text{add 6 to both sides}\\\\2b-6+6\geq9+6-3b\\\\2b\geq15-3b\qquad\text{add}\ 3b\ \text{to both sides}\\\\2b+3b\geq15-3b+3b\\\\5b\geq15\qquad\text{divide both sides by 5}\\\\\dfrac{5b}{5}\geq\dfrac{15}{5}\\\\b\geq3$

$,\ \geq-\text{line to the right}\\-\text{o}\text{pen circle}\\\leq,\ \geq-\text{closed circle}$

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3 years ago