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Ann [662]
4 years ago
5

Jessica and Alex use ribbon to decorate picture frames. Jessica has 1/2 yards of ribbon : She uses ribbon to decorate 3 picture

frames : She uses and equal amount ribbon to decorate each picture frame how many yards of ribbon does Jessica use to decorate each pucture frame A. 1/6 B. 1 C. 3/2 D. 6
Mathematics
1 answer:
iVinArrow [24]4 years ago
7 0

Hello there!

If Jessica uses 3 equal sizes for all three picture frames, and she only has 1/2 of a yard, one easier way that we can convert this would be that there are 0.5 = 18 inches of Ribbon.

18 ÷ 3 = 6 ... 6 × 3 = 18.

Your correct answer would be D. 6

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Sarah works ar a shoe store. She earns a base salary of $1,500 per month, plus a 25% commission on the amount of shoe sales she
STatiana [176]

Answer:

Step-by-step explanation:

goal is 2,000 and she already makes 1,500

2,000-1,500 gives you the amount she needs to earn from shoe sales

she needs to earn 500 dollars with a 25% commision

25% can also be written as 25/100

for every $100 she makes, she gets $25 but she needs $500

100:25

we need the second number in the ratio to equal $500

500 divided by 25 is 20

20 x 100 : 25 x 20

2000: 500

ANOTHER WAY: 500 is 1/4 of 2000 and 1/4 is 25% so you could multiply the amount she needed by 4 to get your answer

3 0
3 years ago
Write the fraction 7/10 as a sum of fraction three different ways
LUCKY_DIMON [66]
A . 2/10 + 5/10= 7/10
B. 3/10 + 4/10= 7/10
C. 6/10 + 1/10 =7/10
4 0
4 years ago
Is 87 composite or prime
ankoles [38]
87 is a composite number 
5 0
4 years ago
Read 2 more answers
A boy has twice as many nickels as he has pennies, and he has 5 less dimes than nickels. If he has $4.46 in all, how many coins
Zarrin [17]
N=2p
n/2=p

n-5=d

10d+5n+1p=446
subsitute n/2 for p
subsitute n-5 for d
10(n-5)+5n+n/2=446
times 2 both sides
20(n-5)+10n+n=892
expand
20n-100+10n+n=892
31n-100=892
add 100 both sides
31n=992
divide both sides by 31
n=32

subsitute back

n/2=p
32/2=16=p


n-5=d
32-5=d=27


27 dimes
16 pennies
32 nickles
7 0
3 years ago
What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?
Akimi4 [234]
\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------

\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5
4 0
3 years ago
Read 2 more answers
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