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Fed [463]
3 years ago
7

Last year, Scout Troop # 85 sold 540 boxes of cookies. This year, they sold 783 boxes. What was the percent increase in the numb

er of boxes sold?
Mathematics
1 answer:
lesantik [10]3 years ago
4 0
Difference/original
783 - 540 = 243
243/540 = 0.45
0.45 * 100 = 45
45% = increased percentage
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R−5r−6s+8s−15 what is the answer for this equation
weeeeeb [17]

Step-by-step explanation:

r-5r-6s+8s-15

-4r+2s-15

= 2s-4r-15

4 0
3 years ago
ZARA'S DAD IS 6 TIMES OLDER THAN ZARA. THE PRODUCT THEIR AGE IS 150 YEARS. CALCULATE THEIR RESPECTIVE AGES
Masteriza [31]

Answer:

Zara = 5
Her Dad = 30

Step-by-step explanation:

Lets say Zara's age is "z"
So, her dad's age must be 6 x z = 6z
As its given that there product is 150 so we can also write it as :

z x 6z = 150
6z to the power 2 = 150
z to the power 2 = 150 divided by 6
                            = 25
So, z = 5
So Zara's age is 5 so her dad's age must be 30

3 0
2 years ago
The basketball team is traveling 347 miles from Nashville, Tennessee to Columbus, Ohio for a national tournament. The team plans
Andrej [43]
I believe the answer is about 3:30 pm.
8 0
3 years ago
Will an object with a density of. 97 g/ml float or sink in water explain
Ainat [17]
I would say the object will float because the density of water is 999.99kg/m which is slightly less than 1g but still more than the object.
3 0
3 years ago
Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
Veronika [31]

The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

\int\limits^5_1 {x/(2+x^{3}) } \, dx=f(x)=x/2+x^{3}

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

=4\lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3}

Hence the expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

#SPJ4

5 0
2 years ago
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