All categories

2 years ago

What is the output of the function when the input is 0?

Mathematics

1 answer:

ohaa [14]2 years ago

7 0

Answer:

The input is zero, the output is -1

Step-by-step explanation:

We want to know the y value when the x value is zero

When x =0, y = -1

The input is zero, the output is -1

You might be interested in

(−m <br> 2<br> +6)+(−4m <br> 2<br> +7m+2)

dmitriy555 [2]

Following are calculation to the given eqution:

Given:

$\bold{(-m^2+6)+(-4m^2+7m+2)}$

To find:

Solve the equation.

Solution:

$\to \bold{(-m^2+6)+(-4m^2+7m+2)}\\\\\to \bold{-m^2+6-4m^2+7m+2}\\\\\to \bold{-5m^2+7m+8}\\\\\to \bold{-1(5m^2-7m-8)}\\\\$

Therefore, the final solution is " $-1(5 m^2-7m-8)$ ".

Learn more:

brainly.com/question/12685776

7 0

2 years ago

What is the domain of the function below? (0,2 (3,1) (5,2) (8,4)

KiRa [710]

{0,3,5,8} ok this0 is the an0sw0er

7 0

3 years ago

Read 2 more answers

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

2 years ago

I need some help with zee maths.

seraphim [82]

You have to collect like terms that means group the numbers that have the same variables

8 0

2 years ago

Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the eve

riadik2000 [5.3K]

Answer:

0.6708 or 67.08%

Step-by-step explanation:

Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:

$P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708$

There is a 0.6708 probability that Helen makes both free throws.

5 0

3 years ago