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4 years ago

Y=5x-25 , the output is -25 what’s the input

2 answers:

8 0

In the equation y = 5x - 25, x is the input and y is the output. You know this because you start with some x value, apply some operations, and you get the answer y. So if -25 is the output, then y=-25. To find the input, just solve for x:
y=5x-25
-25=5x-25
0=5x
x=0
So the input is 0.

Nikitich [7]4 years ago

5 0

Y = 5x - 25

output = -25

-25 = 5x - 25

5x = -25 + 25

5x = 0

x = 0

Answer: Input = 0

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-2 and 1/3 times 1 and 1/2=

torisob [31]

Answer:

$-2\frac{1}{3}\times 1\frac{1}{2}=-3.5$

Step-by-step explanation:

We are asked to find the product of $-2\frac{1}{3}$ and $1\frac{1}{2}$ .

Upon writing our expression as a product we will get,

$-2\frac{1}{3}\times 1\frac{1}{2}$

Let us convert our given mixed fractions to improper fractions.

$\frac{-7}{3}\times \frac{3}{2}$

Upon cancelling out 3 from our given expression we will get,

$\frac{-7}{2}$

Upon dividing -7 by 2 we will get,

$\frac{-7}{2}=-3.5$

Therefore, our expression simplifies to -3.5.

4 0

3 years ago

I dont know how to do this <br>6x-2=4x+2

Thepotemich [5.8K]

Answer:

x=2

Step-by-step explanation:

6x-2=4x+2

Move all terms with x to the left side by subtracting 4x from both sides:

2x-2=2

Move all terms without x to the right side by adding 2 to both sides:

2x=4

Divide both sides by 2 to get x by itself:

x=2

8 0

3 years ago

Find the missing values in the ratio table. Then write the equivalent ratios.

andre [41]

Answer:

8:6, 16:12, 24:18

Step-by-step explanation:

7 0

2 years ago

Please help i’ll give brainless

Mars2501 [29]

Answer:

i think the answer is the fourth opition slope = -5/3

6 0

4 years ago

Read 2 more answers

Help me please???<br> I don’t know how to get started!

Sliva [168]

A differentiable function $f(x)$ is increasing on an open interval $(a,b)$ if $f'(x)>0$ for all $a$ , and decreasing if $f'(x)$ . For this problem, you then need to compute the derivative:

$f(x)=x^2\ln x\implies f'(x)=2x\ln x+x=(2\ln x+1)x$

then solve for $f'(x)=0$ :

$(2\ln x+1)x=0\implies x=0\text{ or }x=e^{-1/2}$

We can ignore $x=0$ because $x^2\ln x$ is defined only for $x>0$ . So we have two intervals to consider, $(0,e^{-1/2})$ and $(e^{-1/2},\infty)$ . All we need to do is pick any value from either interval and check the sign of the derivative $f'(x)$ . Since $e^{-1/2}\approx0.606$ , from the first interval we can take $x=\dfrac12$ , and from the second we can pick $x=1$ .

$f'\left(\dfrac12\right)\approx-0.193$

$f'(1)=1>0$

The above indicates that $f(x)$ is decreasing on the first interval $(0,e^{-1/2})$ , and increasing on the second interval $(e^{-1/2},\infty)$ .

For part (b), we use the info from above as part of the first derivative test for extrema. We have one critical point at $x=e^{-1/2}$ , and we know how $f(x)$ behaves to either side of this point; $f(x)$ decreases to left of it, and increases to the right. This pattern is indicative of a minimum occurring at $x=e^{-1/2}$ , and we find that $f(x)$ has the (local) minimum value of $f(e^{-1/2})=-\dfrac1{2e}\approx-0.184$ .

8 0

3 years ago