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4 years ago

Please help,Thanks;)

Mathematics

1 answer:

vichka [17]4 years ago

4 0

To solve this, you divide the price by the amount of ounces, and find the one which results in the smallest number. That was B, $4.29 for 12 ounces.

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It looks easy i just dont know how to do it can someone help me out

Arada [10]

Answer:

Step-by-step explanation:

16+4x = 10 +14

-16 -16

4x = 8

/4 /4

x=2

8x=8(2)

=16

3 0

3 years ago

Find the volume of a cylinder that has a radius of 11 m and a height of 9 m.

hjlf

Answer:

The volume would be 3,421.19m

7 0

3 years ago

Which number should each side of the equation 3/4 = 9 be multiplied by to produce the equivalent equation of x = 12?

Tema [17]

You need to multiply both sides of the equation by the reciprocal of 3/4. The reciprocal is 4/3 and 4/3 multiplied by 9 = 12. So x = 12 because a number times the reciprocal is one.

6 0

3 years ago

Read 2 more answers

2(3 x5-8)-4x2 (8-6+2)

4vir4ik [10]

Answer:

-18

Step-by-step explanation:

4 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

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

4 0

3 years ago