Jeff provides photos for two online sites: site A and site B.

Mathematics

2 answers:

Romashka-Z-Leto [24]2 years ago

6 0

Answer: $0.25

Step-by-step explanation:

Given: The amount paid by Site A for every photo Jeff provides = $0.65

Then, the amount paid by Site A for 5 photos Jeff provides= $5\times0.65=\$3.25$

Also, the amount in dollars (y) site B pays as a function of the number of photos provided (x) is represented by the equation $y = 0.60x$

So, for x = 5 photos, the amount paid by site B = $y=0.06\times5=\$3.00$

Now, the difference in payment done by Site A and Site B = $\$3.25-\$3.00=\$0.25$

Hence, Jeff was paid $0.25 more at site A than at site B, if he provided five photos for each site.

Setler [38]2 years ago

5 0

$0.25 you just have to multiply .65 times 5 for site A which is $3.25 and for sure b just plug in 5 for x and multiply by .60 which gives you $3.00 so site A gives you $0.25 more

You might be interested in

(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28

Veronika [31]

$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}$