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ella [17]
3 years ago
15

Kray sells personalized photo frames. She charges $12 for the frame plus $0.25 for each letter the customer wants to engraved on

the frame. Which function represents the relationship between the number of letters engraved on a frame, n, and the total cost of a frame, c,? 15 points
Mathematics
1 answer:
VARVARA [1.3K]3 years ago
6 0

Answer:

c=0.25n+12

Step-by-step explanation:

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Danielle earns a 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary
sashaice [31]

Danielle earned $951.25

Step-by-step explanation:

Multiply her commission by her sales and add her base rate

x=7.25%($4500)+625 or x=0.0725(4500)+625

distribute

x=326.25+625

simplify

x=951.25

8 0
3 years ago
adhiambo buys t litres of milk everyday.A litre of milk is she 25.She spends sh 3000 on milk on November.How many litres does sh
Sergeeva-Olga [200]
3000/30=100every day

100/25=4
So she buys 4 L of milk every day.
So t=4.

4 0
3 years ago
A triangle on a coordinate plane is translated according to the rule T–8, 4(x, y). Which is another way to write this rule? (x,
Lapatulllka [165]

Answer:

C (x, y) - (x - 8, y + 4)

5 0
3 years ago
Read 2 more answers
How are the two functions f(x) = 0.7(6)^x and g(x) = 0.7^-x related to each other?
lubasha [3.4K]
To find the relationship between them, lets do some algebra:

f(x) = (7/10)^x

g(x) = (7/10)^-x

if we do:

f(x) * g(x) = 1

due to exponents multiplying addition law, therefore, the relationship, gives:

f(x) = 1 / g(x)
5 0
3 years ago
Refer to attachment for question
AveGali [126]

Answer:

Below!

Step-by-step explanation:

Let the <u>irrational number</u> be known as "x".

\implies -\sqrt{41} \times x = 1

Divide both sides by -√41.

\implies \dfrac{(-\sqrt{41}  \times x)}{-\sqrt{41} }  = \dfrac{1}{\sqrt{-41} }

\implies  x = \dfrac{1}{-\sqrt{41} }

Take the "-" to the <u>numerator:</u>

\implies x = \dfrac{-1}{\sqrt{41} }

Use parenthesis to isolate the "-"

\implies x = -\huge\text{(}\dfrac{1}{\sqrt{41} } \huge\text{)}

Note: 1 can also be written as √1. (√1 = √1 × √1 = 1)

\implies {x = -\huge\text{(}\dfrac{\sqrt{1}}{\sqrt{41} }\huge\text{)}}

Combine the roots in the equation:

\implies {x = -\huge\text{(}\sqrt{\dfrac{1}{41} }} \huge\text{)}}

Remove the parenthesis:

\implies {x = -\sqrt{\dfrac{1}{41} }}

Thus, Option C is correct.

6 0
2 years ago
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