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AlekseyPX
2 years ago
11

You by x packs of pencils, twice as many packs of erasers, and three times as many rolls of tape. Write an expression in the sim

plest form for the total amount of money you spent.
Mathematics
1 answer:
ra1l [238]2 years ago
7 0
Answer- since you are buying x packs of pencils, 2x packs of erasers, and 3x rolls of tape, you would be buying a total of 6x items.
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What is the length of the diagonal of the square shown below?
Usimov [2.4K]
Hey,here is the answer to ur question......!!!!
Since it is a square and all angles =90
Therefore to
Find length of the diagonal we have to use Pythagoras theorem....!
So, (AC)^2=(AB)^2+(BC)^2
=>(AC)^2=(11)^2+(11)^2
=>(AC)^2=121+121
=>(AC)^2=242
=>AC=root242
=>AC=11root2
=>length of the diagonal =11root2
Hope this helps u....!!!!
6 0
2 years ago
Read 2 more answers
1)Given: AB = 4 AD = 6
Vaselesa [24]
<span>1)Given: AB = 4 AD = 6
What is the name of the radius of the larger circle?
the answer part 1) is 
the radius </span>of the larger circle is AD

<span>2)Given: AB = 4 AD = 6
What point is in the interior of both circles?
the answer Part 2) is 
The point A  (the center of the circles)

</span><span>3) Given: AB = 4 AD = 6
Which points are in the exterior of both circles?</span><span>
the answer Part 3) is
</span><span>E and G

</span><span>4)The circles are _____.
</span><span>the answer Part 4) is
</span><span>concentric
</span>
<span>5)If AC = 20 and BD = 8, what is the radius of the smaller circle?
</span>we know that
radius smaller circle=AB
and
AB=AC-BD--------> AC=20-12-------> AC=8 units

the answer part 5) is
the radius of the smaller circle is 12 units

<span>6)Given: AB = 4 AD= 6
What is the length of BD?</span>
we know that
AD=AB+BD
solve for BD
BD=AD-AB--------> BD=6-4-----> BD=2 units

the answer Part 6) is
the length of BD is 2

<span>7)Given: AB = 4 AD = 6
What is the name of the radius of the smaller circle?</span>

the answer Part 7) is 
the name of the radius of the smaller circle is AB
5 0
2 years ago
Kayleigh babysat for 11 hours this week. That was 5 fewer than 2/3 as many hours as she babysat last week, h. Write an equation
lukranit [14]

Answer: The answer is \textup{x}=\dfrac{2}{3}\textup{h}-5.


Step-by-step explanation: Given that Kayleigh babysat for 11 hours the present week. Also, this was 5 less than two-third of the number of hours she babysat last week, which is represented by 'h'.

We are to write an equation to represent the number of hours she babysat each week.

So, for that, let 'x' be the number of hours she babysat this week. Then, according to the question, we can write

\textup{x}=\dfrac{2}{3}\textup{h}-5.

Also, it is given that

\textup{x}=11.

Therefore,

11=\dfrac{2}{3}\textup{h}-5\\\\\Rightarrow \dfrac{2}{3}\textup{h}=16\\\\\Rightarrow \textup{h}=24.

Hence, using the above relation, we can find the number oh hours Keyleigh babysat each week.

Thus, the required equation is

\textup{x}=\dfrac{2}{3}\textup{h}-5,

where, 'x' and 'h' are the number of hours she sat this week and last week respectively.

 


4 0
2 years ago
What the number is 1/10 the value of 237
Troyanec [42]
1/10 the value of 237 means we must multiply 1/10 by 237

1/10 * 237 = 23.7

Hope this helps!
4 0
2 years ago
An astronaut on the moon throws a baseball upward. the astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is
marta [7]

Answer:

Step-by-step explanation:

The position function is

s(t)=-2.7t^2+50t+6.5 and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:

10=-2.7t^2+50t+6.5 and

0=-2.7t^2+50t-3.5 and factor that however you are currently factoring quadratics in class to get

t = .07 sec and t = 18.45 sec

There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).

For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:

0=-2.7t^2+50t+6.5 and factor that to get

t = -.129 sec and t = 18.65 sec

Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.

4 0
1 year ago
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