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Black_prince [1.1K]
2 years ago
8

Li wants to add these items to her suitcase: a hairdryer (1.25 lb), a hand-held video game (0.6 lb), an extra video game (0.25 l

b), and a music player (0.4 lb). Using the inequality x ≤ 2.25, can she add all these items to the suitcase? If not, what could be left out to be closest to the maximum weight allowed? Explain.
Her suitcase already is 47.75 with a 50 lbs limit. There is an extra fee once you go past the limit

PLZ HELP!!
Mathematics
2 answers:
lbvjy [14]2 years ago
7 0
Add some of them or all of them to your sum of 47.75, if either or exceeds the limit then that is what left out.
Rzqust [24]2 years ago
6 0

The total weight of the items is 1.25 + 0.25 + 0.6 + 0.4 = 2.5 lb. 2.5 is 0.25 greater than 2.25, so not all the items can be added. If Li leaves out the extra game, which weighs 0.25 lb, she’d have the greatest possible weight allowed.

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On a trip, you had to change your money from dollars to euros. You got 450 euros for 600 dollars. Which is a unit rate that desc
chubhunter [2.5K]
The number of Euros received against 600 dollars = 450
Then
For 1 dollar, the number of euros = (450/600) euros
                                                       = (45/60) euros
                                                       = ( 3/4) euros
                                                       = 0.75 euros
So the unit rate that describes the exchange rate is 0.75 euros per dollar. So the correct option among all the options given in the question is option "A". I hope the procedure for getting to the correct result is clear enough for you to understand. In future you can use this procedure for solving similar problems.


7 0
2 years ago
Write g(x) = 4x2 + 88x in vertex form. The function written in vertex form is g(x) = ___ (x +11)2 +____.
wel

The\ vertex\ form:\\\\f(x)=a(x-h)^2+k

\text{Use}\ (a+b)^2=a^2+2ab+b^2\qquad(*)

g(x)=4x^2+88x=4(x^2+22x)=4(x^2+2\cdot x\cdot11)\\\\=4(\underbrace{x^2+2\cdot x\cdot11+11^2}_{(*)}-11^2)=4[(x+11)^2-121]=4(x+11)^2+(4)(-121)=4(x+11)^2-484

5 0
1 year ago
Read 2 more answers
For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
1 year ago
Read 2 more answers
Emily is thinking of a number, which she calls n. She finds 1/4 of the number and then subtracts 3. Write an expression to repre
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Answer:

1/4n-3

Step-by-step explanation:

3 0
1 year ago
The height of a baseball thrown from the catcher to first base is modeled by the function h(t) = –0.09t2 + 0.72t + 6, where h is
Black_prince [1.1K]

The height of a baseball thrown from the catcher to first base is modeled by the function h(t) = -0.09t^2 + 0.72t + 6, where h is the height of the ball measured in feet and t is the time since being thrown, measured in seconds

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h(t) = -0.09t^2 + 0.72t + 6

h(10) = -0.09(10)^2 + 0.72(10) + 6

h(10)=4.2 feet

the height of the ball at 10 seconds= 4.2 feet

3 0
1 year ago
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