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Nina [5.8K]
1 year ago
5

Sally wants to purchase D dozen donuts for $4.50 each and B boxes of cookies for $2.25 each. She can spend at most $30. Which in

equality could be used to determine how many of each item Sally can purchase?
a
4.50D + 2.25B ≤ 30

b
4.50D + 2.25B < 30

c
4.50D + 2.25B > 30

d
4.50D + 2.25B ≥ 30
Mathematics
1 answer:
maw [93]1 year ago
5 0

Considering the definition of an inequality, the inequality that could be used to determine how many of each item Sally can purchase is 4.50D + 2.25B ≤ 30.

<h3>Definition of inequality</h3>

An inequality is the existing inequality between two algebraic expressions, connected through the signs:

  • greater than >.
  • less than <.
  • less than or equal to ≤.
  • greater than or equal to ≥.

An inequality contains one or more unknown values ​​called unknowns, in addition to certain known data.

Solving an inequality consists of finding all the values ​​of the unknown for which the inequality relation holds.

<h3>Inequality in this case</h3>

In this case, you know:

  • Sally wants to purchase D dozen donuts for $4.50 each.
  • Sally wants to purchase B boxes of cookies for $2.25 each.
  • She can spend at most $30.

The inequality that expresses the previous relationship is

4.50D + 2.25B ≤ 30

Finally, the inequality that could be used to determine how many of each item Sally can purchase is 4.50D + 2.25B ≤ 30.

Learn more about inequality:

brainly.com/question/17578702

brainly.com/question/25275758

brainly.com/question/14361489

brainly.com/question/1462764

#SPJ1

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Increased height:-

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3 years ago
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How would you solve (-1,4),y=3x+5
astraxan [27]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
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If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

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8 0
3 years ago
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