Round off 19340 to 1 s f

Mathematics

1 answer:

dimulka [17.4K]3 years ago

5 0

Answer:

20000

Step-by-step explanation:

This is because when you round to the signifigant figure, you basically look at any digits at the start of the number, which isn't zero. In this case it would be in the ten thousand 19340. So you round 19340 to the nearest 10 thounsand, which, would be 20000

You might be interested in

during a softball game kay hit a fly ball the function f(x) = -16t^2 + 64t + 4 describes the height of the softball in feet. mak

STatiana [176]

<h2>Hello!</h2>

The graph is attached.

To graph a parabola we need to know the following:

- If the parabola is open upwards or downwards

- They axis intercepts (if they exist)

- The vertex position (point)

We are given the function:

$f(t)=-16t^{2}+64t+4$

Where,

$a=-16\\b=64\\c=4$

For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.

Finding the axis interception points:

Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:

$f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0$

Using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615$

So, at t=4.0615 the height of the softball will be 0.

Since we will work only with positive values of "x", since we are working with a function of time:

Let's start from "t" equals to 0 to "t" equals to 4.0615.

So, evaluating we have:

$f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0$