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Rudik [331]
2 years ago
9

How many hundreds flats are equal to 400 tens rods

Mathematics
2 answers:
melomori [17]2 years ago
7 0
40 hundred flats . 400 tens = 4,000. 40 hundreds also eqals 4,00
vitfil [10]2 years ago
4 0
40 hundreds flats. 400 tens = 4,000. 40 hundreds also equals 4,000.
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gayaneshka [121]
I would go with d .... If not it would be c...


Really hope I help but I would choose d
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2 years ago
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Which statement best describes how to determine whether f(x) = 9 – 4x2 is an odd function? Determine whether 9 – 4(–x)2 is equiv
katovenus [111]

We have a property for odd functions, which is given below. Let f(x) be an odd function then it must satisfy the below - mentioned property.

f(-x)= -f(x)

Now, we have been given the function f(x)=9-4x^2

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f(-x)=9-4(-x)^2

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Therefore, C is the correct option.

5 0
2 years ago
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The revenue function for a production by a theatre group is R(t) = -50t^2 + 300t where t is the ticket price in dollars. The cos
Mkey [24]
Break even is the value of t where revenue=cost or R(t)=C(t)

set equal each other

-50t^2+300t=600-50t
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divide both sides by 50
t^2-6t=t-12
minus t-12 from both sides
t^2-7t+12=0
factor
(t-4)(t-3)=0
set each to zero
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t-3=0
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7 0
1 year ago
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Two students are using estimation to determine reasonable solutions to the expression 89.1 times 9.3. Katie uses the expression
dmitriy555 [2]
Amaya’s because she rounded to the nearest tenth correctly if the number behind the decimal had been 5 or above Katie would be right but because the number is 4 or below you round down therefore 89.1 would be 89 and 9.3 would be 9
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1 year ago
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Tasha invests in an account that pays 1.5% compound interest annually. She uses the expression P(1+r)t to find the total value o
zhenek [66]
\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$3000\\
r=rate\to 1.5\%\to \frac{1.5}{100}\to &0.015\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &5
\end{cases}
\\\\\\
A=3000\left(1+\frac{0.015}{1}\right)^{1\cdot 5}\implies A=3000(1.015)^5\implies A\approx 3231.852
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1 year ago
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