Stephanie put $80 in her bank account when she was five years old. The bank gave her a simple interest rate of 2.1%.

Kaitlyn gave 1/2 of her candy bar to arianna. arianna gave 1/3 of the candy she got from kaitlyn to cameron. what fraction of a

sukhopar [10]

Let the total candy bar be represented by x.

Amount of chocolate Kaitlyn gave to Arianna = $\frac{1}{2}x$

Amount of chocolate Arianna gave Cameron

= $\frac{1}{3}$ rd <span>of the candy she got from Kaitlyn to Cameron

= </span> $\frac{1}{3}$ × $\frac{1}{2}x$

= $\frac{1}{3}* \frac{1}{2} x = \frac{1}{6} x$

Hence, Cameron got $\frac{1}{6}$ th of the original chocolate.

5 0

1 year ago

Is the statement "Two matrices are row equivalent if they have the same number of rows" true or false? Explain. A. True, because

vlabodo [156]

Answer: The answer is (C).

Step-by-step explanation: The given statement is - "Two matrices are row equivalent if they have the same number of rows". We are to explain whether the statement is true or false.

What are row equivalent matrices? The answer to this question is -

Two matrices are said to be row equivalent if one of the matrices can be obtained from the other by applying a number of elementary row operations. Or, we can say two matrices of same order are row equivalent if they have same row space.

Thus, the correct option is (C).

7 0

2 years ago

Ms. Turner drove 825 miles in March. She drove 3 times as many miles in March as she did in January. She drove 4 times as many m

navik [9.2K]

825/3 = 275 which is how many miles she drove in January.

275*4= 1100 which is how many miles she drove in February.

Ms. Turner drove 1100 miles in February.

4 0

2 years ago

What is a3 in an arithmetic sequence in which a10=41 and a15=61

USPshnik [31]

$\bf \begin{array}{llll}
term&value\\
-----&-----\\
a_{10}&41\\
a_{11}&41+d\\
a_{12}&(41+d)+d\\
&41+2d\\
a_{13}&(41+2d)+d\\
&41+3d\\
a_{14}&(41+3d)+d\\
&41+4d\\
a_{15}&(41+4d)+d\\
&41+5d=61
\end{array}
\\\\\\
41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\
-------------------------------\\\\$

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=10\\
a_{10}=41
\end{cases}
\\\\\\
41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}$

thus

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
d=4\\
n=3\\
a_{1}=5
\end{cases}
\\\\\\
a_3=a_1+(3-1)4\implies a_3=5+(3-1)4$

and surely you know how much that is.

8 0

2 years ago

Read 2 more answers

A particular beach is eroding at a rate of 4 centimeters per year. A realtor converts this rate to millimeters per day. Which ex

arsen [322]

4 cm/year * 10mm/cm * 1year/365day ≈ 0.11 mm/day

8 0

1 year ago

Read 2 more answers