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3 years ago

Round 224.91 to the nearest whole number do not add extra zeros

Mathematics

2 answers:

-BARSIC- [3]3 years ago

5 0

Answer:

225

Step-by-step explanation:

Bogdan [553]3 years ago

5 0

Answer:

Step-by-step explanation:

I'm pretty sure 224.91 is closest to 225.

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F(x) =2x-5 and g(x) =x+52 find f(g(x)

irga5000 [103]

If f(x) = 2x - 5 and g(x) = x + 52, then f(g(x)) can be deduced by placing g(x) in the spot of x in the f(x) equation as follows:

f(g(x)) = 2(g(x)) - 5

Since we know g(x) = x + 52, let's plug it in:

f(g(x)) = 2(x + 52) - 5
f(g(x)) = 2x + 104 - 5
f(g(x)) = 2x + 99

4 0

3 years ago

Read 2 more answers

a boy is standing on a pole of height 14.7m throws a stone upwards. it moves in a vertical line slightly away from the pole and

fomenos

Answer:

The time taken for the upward motion is 1 second. The same time is taken for the downward motion

It reaches a maximum height of 4.9 meters.

Step-by-step explanation:

The equation of motion is:

$x(t) = -4.9t^{2} + 9.8t$

Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.

Vertex of a quadratic equation:

Quadratic equation in the format $x(t) = at^{2} + bt + c$

The vertex is the point $(t_{v}, x(t_{v}))$ , in which

$t_{v} = -\frac{b}{2a}$

In this question:

$x(t) = -4.9t^{2} + 9.8t$

So $a = -4.9, b = 9.8$

Vertex:

$t_{v} = -\frac{9.8}{2*(-4.9)} = 1$

The time taken for the upward motion is 1 second.

$x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9$

It reaches a maximum height of 4.9 meters.

Downward motion:

From the vertex to the ground.

The ground is t when x = 0. So

$-4.9t^{2} + 9.8t = 0$

$4.9t^{2} - 9.8t = 0$

$4.9t(t - 2) = 0$

$4.9t = 0$

$t = 0$

$t - 2 = 0$

$t = 2$

It reaches the ground when t = 2 seconds.

The downward motion started at the vertex, when t = 1.

So the duration of the downward motion is 2 - 1 = 1 second.

5 0

3 years ago

The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func

alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

$h(t)=-16t^2 + 96t$

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$