Evaluate 7 + (-4x2) for x = 0.

Mathematics

2 answers:

krek1111 [17]3 years ago

7 0

-4 × 0^2 = 0
7 + 0 = 7

aksik [14]3 years ago

5 0

Answer: The required value of the expression at x = 0 is 7.

Step-by-step explanation: We are given to evaluate the following expression for x = 0.

$g(x)=7+(-4x^2)~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

To evaluate, we must substitute x = 0 in the given expression.

So, putting x = 0 in equation (i), we get

$g(x)\\\\=7+(-4x^2)\\\\=7-4x^2\\\\=7-4\times 0^2\\\\=7-0\\\\=7.$

Thus, the required value of the expression at x = 0 is 7.

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OleMash [197]

Answer:

area = 22,272

Step-by-step explanation:

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$87^{2} + x^{2} = 145^{2}$

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4 0

3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below