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2 years ago

Explain the difference between evaluating 3 * 7 - 4 ÷ 2 and evaluating 3(7 - 4) ÷ 2.

Mathematics

2 answers:

Bogdan [553]2 years ago

6 0

DIY. I believe in you, its just a matter if you believe in yourself. Your welcome

shusha [124]2 years ago

5 0

In 3(7-4)/2 you have you have to multiply 3 by 7 and by -4 and the answer would be 21-12/2=9/2 or 4.5. While in 3 * 7 -4/2 you will have to multiply 3 by 7 only and your answer will be 21 -4/2=17/2 or 8.5.

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Give 6 irrational numbers between 2÷5 and 3÷4

soldi70 [24.7K]

Hello,

Let's z=0.1234567891011121314151617181920.....(never end)

2/5=(2/(5z)*z<3241/1000 *z
3/4=(3/(4z))*z>6075/1000*z

So here is 2836 irrational numbers:

3241*z/1000
3242*z/1000
3243*z/1000
...
6073*z/1000
6074*z/1000
6075*z/1000

7 0

2 years ago

Wes bought a conference table for $960. what is it worth after depreciating at a rate of 12% per year for 4 years?

murzikaleks [220]

Answer:

$594.03

Step-by-step explanation:

Using the exponential function;

A = Pe^-rt

Principal = $960

Rate r = 12% = 0.12

Time t = 4years

Substitute

A = 960e^-(0.12*)*4

A = 960e^-0.48

A = 960(0.61878)

A = 594.03

HEnce the worth after 4years will be $594.03

4 0

2 years ago

Is this right???????? Please help

Lady bird [3.3K]

It would be 8 bro dududuudududud

7 0

3 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

2 years ago

Two integers from 1 through 40 are chosen by a random number generator. write your answer as a fraction in simplest form a/b. p(

jasenka [17]

Probability is 1/40.

5 0

3 years ago