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son4ous [18]
3 years ago
11

So if something thats $4 for 10k how much would $2 be? 5k? 3k?

Mathematics
1 answer:
Stells [14]3 years ago
5 0

Answer:

5k

Step-by-step explanation:

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A classroom had 8 rows of desks with 6 desks in each row. Four desks were removed from the room. How many desks were left?
Oduvanchick [21]
Answer: 44 desks.

First of all, you need to find how many desks there are in total. To find it, you need to multiply 8 with 6.

8 x 6 = 48

Next, it says that four desks were removed. All you need to do is subtract 4 from 48.

48 - 4 = 44.
5 0
3 years ago
681/4 i dont know this
ankoles [38]

Answer:

170.25

Step-by-step explanation:

It's a bit tricky to do not on plain paper, but here:

Just assume the long division symbol looks right

  _1_7_0_.25

4 ) 681

    -4

     2 8

     -2 8

          0  1

The remainder is "1", so you'll just have to write that as either a fraction or decimal, whatever is assigned.

Fraction: 170 1/4

Decimal: 170.25

7 0
3 years ago
Read 2 more answers
A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20 degrees above the ground from a height of 4.5 ft.
Leviafan [203]

Answer:

T= 2.35 seconds

Step-by-step explanation:

⇒The question is on the time of flight.

⇒Time of flight is the time taken for a projected object to reach the ground.It depends on the <u>projectile angle</u> and the <u>initial velocity</u> of the projectile

Given;

Initial velocity of ball= 110ft./sec.

The projectile angle= 20°

Acceleration due to gravity, g=32 ft./s²

⇒Formulae for time of fright T= (2×u×sin Ф)/g

Where T=time of fright, u=initial velocity of projectile, Ф=projectile angle and g=acceleration due o gravity.

<u>Substituting values</u>

T= (2×u×sin Ф)/g

T=( 2×110×sin 20°) / 32

T= 2.35 seconds

3 0
3 years ago
Read 2 more answers
Someone can help?? ​
oee [108]

Answer:

Step-by-step explanation:

let  cos^{-1}x=t

cos t=x

when x=1,cos t=1=cos 0

t \rightarrow 0

\lim_{x \to 1} \frac{1-\sqrt{x}}{(cos ^{-1}x)^2 } \\= \lim_{t \to 0} \frac{1-\sqrt{cos~t}}{t^2} \times \frac{1+\sqrt{cos~t}}{1+\sqrt{cos ~t}} \\= \lim_{t \to 0} \frac{1-cos~t}{t^2(1+\sqrt{cos~t})}}  \\= \lim_{t \to 0 }\frac{2 sin^2~\frac{t}{2}}{t^2(1+\sqrt{cos~t})}} \\= 2\lim_{t \to 0 }(\frac{sin~t/2}{\frac{t}{2} })^2 \times \frac{1}{4} \times  \lim_{t \to 0 }\frac{1}{1+\sqrt{cos~t}} \\=\frac{2}4} \times 1^2 \times \frac{1}{1+\sqrt{cos~0}} \\=\frac{1}{2} \times \frac{1}{1+1} \\=\frac{1}{4}

4 0
4 years ago
I need help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Maksim231197 [3]

Answer:

c) I think'

Step-by-step explanation:

7 0
3 years ago
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