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Natali5045456 [20]

2 years ago

6

Help help help help help

2 answers:

baherus [9]2 years ago

5 0

${4}^{5} \div {4}^{2} \\ = {4}^{5 - 2} \\ = 4^{3}$

Sindrei [870]2 years ago

4 0

4^5/4^2 = 4^3

the answer is 4^3 because when you are dividing with exponents, you subtract them. in this case i subtracted 5 & 2 (5-2=3)

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Write in your own words and understanding how do soil formed and identify the factors that helps in the formation of soil.

Anna35 [415]

Answer:

Soil minerals are the foundation of the soil. They are made from rocks (parent material) that have been weathered and eroded naturally. Water, wind, temperature change, gravity, chemical interaction, live creatures, and pressure changes are all factors that aid in the breakdown of parent material.

Step-by-step explanation:

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6 0

2 years ago

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kondor19780726 [428]

The right answer is Option A: $x=\frac{rs-t}{2}$

Step-by-step explanation:

Given expression is;

$s=\frac{2x+t}{r}$

We will solve this expression for x, therefore,

Multiplying both sides by r

$r*s=\frac{2x+t}{r}*r\\rs=2x+t$

Subtracting t from both sides

$rs-t=2x+t-t\\rs-t=2x\\2x=rs-t$

Dividing both sides by 2

$\frac{2x}{2}=\frac{rs-t}{2}\\x=\frac{rs-t}{2}$

The right answer is Option A: $x=\frac{rs-t}{2}$

Keywords: subtraction, division

Learn more about subtraction at:

#LearnwithBrainly

7 0

3 years ago

The equation x + c = k has a solution of x = 2. Both cand k are rational numbers.

jeka57 [31]

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2 years ago

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What is the value of 7^2-3^3

lbvjy [14]

Answer:

22

Step-by-step explanation:

7^2-3^3

(7)(7)-(3)(3)(3)

49-(3)(9)

49-27

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3 0

2 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago