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Novay_Z [31]
2 years ago
13

At the calendar shop, wall calendars cost $10 and desk calendars cost $6. Kira spent $60 to buy 8 calendars. How many of each ty

pe of calendar did Kira buy?
Mathematics
1 answer:
Natali5045456 [20]2 years ago
8 0

Answer: 3 desk calender's and 3 wall calendars

Step-by-step explanation: 6 x 5 = 30 and 3 x 10 = 30

30 + 30 = 60

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Ok thank you also If you dont mind but please click on my pfp and answer my questions please?

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3 years ago
Plzzzzz help !.!.!.!.!.!.!!..!!.!.!.!.!.!.?
zavuch27 [327]

Answer:

1.8

Step-by-step explanation:

there are 100 cementers in a meter, 1000 millimetres in a meter, 1000/100=10 ,so 18/10=1.8

4 0
3 years ago
Read 2 more answers
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
3 years ago
Is 1/5 a repeating decimal​
Vera_Pavlovna [14]

Answer:

o.2222

Step-by-step explanation:

just divide one by five

hope that helps

8 0
3 years ago
Read 2 more answers
Which inequality represents all possible solutions of -3n &lt; -18?
faust18 [17]
I hope this helps you




-3n÷-3 < -18÷-3



n>6
7 0
3 years ago
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