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3 years ago

Log base 5 (3) - log base 5 (5x) = 2

1 answer:

5 0

The answer is 3/125

$log_a(x) - log_a(y)=log_a( \frac{x}{y} ) \\ 
log_5(3) - log_5(5x) = log_5( \frac{3}{5x})$

$log_5( \frac{3}{5x})=2$

Since $log_b(x)=y$ is $b^{y} =x$ , then:
$log_5( \frac{3}{5x})=2$ is
$5^{2} = \frac{3}{5x} \\ 25 = \frac{3}{5x} \\ 5x*25=3 \\ 125x=3 \\ x=3/125$

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34kurt

Answer:

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Step-by-step explanation:

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6 0

3 years ago

For the graph, what is a reasonable constraint so that the function is at least 300?

blsea [12.9K]

Answer:

C) 0 ≤ x ≤ 25

Step-by-step explanation:

We are supposed to find a reasonable constraint so that the function is at least 300 i.e. the value of x at which f(x) is greater or equal to 300

A)x ≥ 0

Refer the graph

At x = 0

f(x)=300

On increasing the value of x , f(x) increases but at x = 12 it starts decreasing

So, x ≥ 0 can also have f(x)<300

So, Option A is wrong

B)−5 ≤ x ≤ 30

At x = -5

f(x) = 100

So, Option B is wrong since we require f(x) is greater or equal to 300

c)0 ≤ x ≤ 25

At x = 0

f(x)=300

At x = 12 , it starts decreasing

At x = 25

f(x)=300

So, The value of f(x) is at least 300 when 0 ≤ x ≤ 25

D)All real numbers

At x = 30

f(x)=0

But we require f(x) greater or equal to 300

Hence Option C is true

8 0

3 years ago

I need help ASAP please I’m giving brainliest!

Ira Lisetskai [31]

I think it would be 35

4 0

3 years ago

Read 2 more answers

Determine whether or not the vector field is conservative. if it is, find a function f such that f = ∇f. if the vector field is

amm1812

$\dfrac{\partial f}{\partial x}=3y^2z^3\implies f(x,y,z)=3xy^2z^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=6xyz^3+\dfrac{\partial g}{\partial y}=6xyz^3$
$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=9xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}=9xy^2z^2$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$\implies f(x,y,z)=3xy^2z^3+C$

The potential function exists, so $\mathbf f$ is indeed conservative.

7 0

3 years ago

The area of a playground is 20 square yards. The length of the playground is 5 times longer than its width. Find the length and

jok3333 [9.3K]

The answer is <span>length = 10 yd, width = 2 yd
proof
</span><span>The length of the playground is 5 times longer than its width, it means
l=5w, but 10= 2x5 so we can write l= 5w,
and Area = 20 =l x w = (5x2) x 2 =20</span>

6 0

3 years ago