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Rom4ik [11]
3 years ago
7

Explain why “Bending” is considered two complementary forces and not just one force. (Make sure to explain what 2 forces are inv

olved, where they are involved and how they change the material as it bends)
srry this is science
Mathematics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

<h2>There are two forces involves because we need to use two points of application to bend something.</h2>

Step-by-step explanation:

We know that "bending" means to force or shape something into a curve or certain angle.

Now, when we bend something we are actually using a pair of forces, not just one, because you can't bend using only one forces, that would imply using one hand which is impossible. Notice that you need to forces to be applied at different points in the same direction to create the curve, these forces must be applied a certain distance away from the equilibrium point which is between the points of application.

Therefore, there are two forces involves because we need to use two points of application to bend something.

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PLEASE HELP AND BE CORRECT
Sergio [31]

(b⁷/4⁵)⁻³

= (4⁵/b⁷)³

= 4¹⁵/b²¹

Therefore answer is option b

Must click thanks and mark brainliest

3 0
3 years ago
Plz help me well mark brainliest if correct
Ira Lisetskai [31]

Answer:

The answer should be B

6 0
3 years ago
Read 2 more answers
Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
Andre45 [30]

Answer:

Option d)  5 to the power of negative 5 over 6 is correct.

\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}

Above equation can be written as 5 to the power of negative 5 over 6.

ie, 5^\frac{\bf -5}{\bf 6}

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}

Above equation can be written as 5 to the power of negative 5 over 6.

7 0
3 years ago
Recursive formula for 26, 24, 22, 20
dexar [7]

Answer:

a_n = 28-2n

Step-by-step explanation:

Given sequence is:

26,24,22,20

We can see that the difference between consecutive terms is same so the sequence is an arithmetic sequence

The standard formula for arithmetic sequence is:

a_n = a_1+(n-1)d

Here,

a_n is the nth term

a_1 is the first term

and d is the common difference

So,

d = 24-26

= -2

a_1 = 26

Putting the values of d and a_1

a_n = 26 + (n-1)(-2)\\a_n = 26-2n+2\\a_n = 28-2n

Hence, the recursive formula for given sequence is: a_n = 28-2n ..

4 0
2 years ago
P=3g-7/k solve for g(plz show work
cestrela7 [59]

Answer:

Step-by-step explanation:

(3g - 7)/k = p

3g - 7 = pk

3g = pk + 7

g = (pk + 7)/3

8 0
3 years ago
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