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tatiyna
3 years ago
7

In the number 1458346.8 what number is in the tenth place

Mathematics
2 answers:
vlada-n [284]3 years ago
4 0

Answer:

8

Step-by-step explanation:

The first digit after the decimal point is called the tenths place value. There are six tenths in the number O.6495. The second digit tells you how many hundredths there are in the number. The number O.6495 has four hundredths

mrs_skeptik [129]3 years ago
4 0

Answer:

The last number (number 8) is in the tenth's place.

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What is the area of this composite figure?
jek_recluse [69]

Answer:

7.14 in. sq

Step-by-step explanation:

First find the area of the square; 2x2=4

Now find the area of the semi-circle; radius = 1.    

1 squared= 1 Now multiply by 3.14 which equals 3.14

Now divide by 2 because it's a semi-circle, you get 1.57.

Multiply 1.57 by 2 because you have 2 semi-circles, the would equal 3.14

Add 4+3.14 and you get 7.14

The area of the composite figure is 7.14 in. sq

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3 years ago
What is the first step to solve number one??
Romashka-Z-Leto [24]

By PEMDAS (or any alternative), <em>the first step should be to divide</em>, as D comes before A. So, 16 ÷ 4.

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2 years ago
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Elizabeth attempts a field goal by kicking a football from the ground with an initial vertical
sammy [17]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t\implies \stackrel{\textit{hits the ground}~\hfill }{0=-16t^2+64t} \\\\\\ 0=-16t(t-4)\implies t= \begin{cases} 0\\ \boxed{4} \end{cases}

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.

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3 years ago
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Sergio039 [100]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t


now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)


\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)

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2 years ago
How many sections are there when you fold a paper 8 times
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32 if you fold it 4 times and times that by 2 it will be 32

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