What digit is in the tenths place in this number? 748.59

2 answers:

6 0

The tenth place is always the position directly to the right of the decimal point. So, the digit in the tenths place of this example is 5.

Hope this helps

Kamila [148]3 years ago

4 0

5 would be the digit in the tenths place

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30 points!!! need help asap ;;

Dominik [7]

How do u see the picture

Step-by-step explanation:

3 0

2 years ago

ASAAPPPPPP!!!!! Complete the tables of values

velikii [3]

Answer:

see explanation

Step-by-step explanation:

Using the rules of exponents

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$ and $a^{0}$ = 1

Substitute the values of x into $4^{-x}$

x = 0 → $4^{0}$ = 1 ⇒ a = 1

x = 2 → $4^{-2}$ = $\frac{1}{4^{2} }$ = $\frac{1}{16}$ , hence

b = $\frac{1}{16}$

x = 4 → $4^{-4}$ = $\frac{1}{4^{4} }$ = $\frac{1}{256}$ , hence

c = $\frac{1}{256}$

6 0

2 years ago

Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati

Nutka1998 [239]

the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9$

$\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill$