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IRINA_888 [86]
3 years ago
9

22.00+27.51+21.33+22.95=

Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

93.79

Step-by-step explanation:

22.00 + 21.33 = 43.33

22.95 + 27.51 = 50.46

43.33 + 50.46 = 93.79

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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

Or

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
Fill in the blank
Marina86 [1]

Answer:

2/66 is ur answer

Step-by-step explanation:

because it will give u ur answer

6 0
3 years ago
HELP HELP HELP HELP HELP HELP HELP HELP HELP
blsea [12.9K]

-38 due to the fact the lower the number, the colder it is.

4 0
3 years ago
Read 2 more answers
The perimeter of a rectangular table is 8 m.
ipn [44]

Answer:

The side lengths are 1.5m and 2.5m.

Step-by-step explanation:

8 0
3 years ago
How do you solve these problems?
Rzqust [24]
A)

\bf log_4(x)=3\\\\
-----------------------------\\\\
log_{{  a}}{{  a}}^x\implies x\qquad \qquad 
\boxed{{{  a}}^{log_{{  a}}x}=x}\impliedby 
\begin{array}{llll}
\textit{we'll use this}\\
\textit{cancellation rule}
\end{array}\\\\
-----------------------------\\\\
4^{\cfrac{}{}log_4(x)}=4^3\implies x=4^3

b)

\bf 3^{2y}=81\qquad 
\begin{cases}
81=3\cdot 3\cdot 3\cdot 3\\
\qquad 3^4
\end{cases}\implies 3^{2y}=3^4\impliedby 
\begin{array}{llll}
\textit{same base}\\
exponents\\
must\ be\\
the\ same
\end{array}
\\\\\\
2y=4\implies y=\cfrac{4}{2}\implies y=2
6 0
4 years ago
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