All categories

3 years ago

What is the constant of proportionality

Mathematics

1 answer:

nalin [4]3 years ago

5 0

Answer:

Step-by-step explanation:

the graph's line is going up by 4

You might be interested in

If (x, 1/100) lies on the graph of y = 10^x, then x =<br><br> 2<br> -1/2<br> -2

stepladder [879]

TLDR: The answer is -2.

What you have here is a function with a point on the graph. It is understood that “x” represents the input and “y” represents the output, so we’re looking for the right “x” that will give y = 1/100.

To start, we can substitute the y-value into the function like this:

y = 10^x

1/100 = 10^x

From here, we can take two different pathways; one requires conceptual knowledge while the other relies on the knowledge of logarithms.

Conceptual

For the conceptual, we know that 100 = 10^2, so 1/100 is the same as 1/10^2. We also know that the inverse of a fraction is the same as the number to the negative of its power (for example, 1/2 is equal to 2^-1), so we know that 1/10^2 is the same as 10^-2. So far, we have learned that:

1/100 = 10^x

1/10^2 = 10^x

10^-2 = 10^x

So, to satisfy this, “x” must equal -2.

Logarithms

For logarithms, we can use powers and math to actually calculate the value of “x”. We know that:

1/100 = 10^x

To solve this, we need “x” to be by itself on one side of the equation. To do this, we can perform the inverse function of an exponent, which is a logarithm. In this case, the base of 10^x is 10, so we need a “base 10” logarithm to solve this function. Apply this function to both sides and simplify:

1/100 = 10^x

log10(1/100) = log10(10^x)

log10(1/100) = x

The log10 function is the inverse of 10^x, so they cancel out to leave “x”. Plug log10(1/100) into a calculator, and you find that x = -2.

Hope this helps!

8 0

3 years ago

What is wrong with the equation? 2 x−3 dx = x−2 −2 2 −3 = − 5 72 −3 f(x) = x−3 is continuous on the interval [−3, 2] so FTC2 can

Wewaii [24]

Answer:

Hello your question is incomplete below is the complete question

What is wrong with the equation? integral^2 _3 x^-3 dx = x^-2/-2]^2 _3 = -5/72 f(x) = x^-3 is continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous at x = -3, so FTC2 cannot be applied The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. If f(2) = 14, f' is continuous, and f'(x) dx = 15, what is the value of f(7)? F(7) =

answer : The value of f(7) = 29

Step-by-step explanation:

Attached below is the detailed solution

Hence : F(7) - 14 = 15

F(7) = 15 + 14 = 29

8 0

4 years ago

(1/4)^x= 256

iren [92.7K]

X= -4
Work: 1^x/4^x=256
1^x= 256 * 4^x
In 1^x = In 256 * 4^x
xIn1 = In 256+ In4^x
x*0= xIn 4 + In 2^8
0= xIn 2^2 + 8In 2
x In 2^2 + 8In 2 =0
x* 2In 2 + 8In 2=0
2In 2x + 8In 2=0
2In 2x = -8In 2
x= -8/2
x=-4

7 0

3 years ago

Pls help on this one?

Gennadij [26K]

Answer:

$\large\boxed{D.\ x+4}$

Step-by-step explanation:

$\dfrac{x^2-16}{x-4}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2}{x-4}=\dfrac{(x-4)(x+4)}{x-4}\\\\\text{canceled}\ (x-4)\\\\=\dfrac{x+4}{1}=x+4$

3 0

4 years ago

PLS HELP

FrozenT [24]

They are similar, but I don't exactly know why. It has something to do with the angles, I know that.

3 0

3 years ago

Read 2 more answers