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3 years ago

Find x. 20 cm 12 cm Х X = cm

Mathematics

1 answer:

frez [133]3 years ago

6 0

Use pythagorean therom, so A^2 would be 12 since it’s the same lengths the square sides, and C^2 is 20. X= 16

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Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago

Solve the equation 4x^2-12x=7 algebraically for x.

ss7ja [257]

  4x² - 12x = 7
  - 7 - 7
4x² - 12x - 7 = 0
  4x² + 2x - 14x - 7 = 0
2x(2x) + 2x(1) - 7(2x) - 7(1) = 0
  2x(2x + 1) - 7(2x + 1) = 0
  (2x - 7)(2x + 1) = 0
2x - 7 = 0 or 2x + 1 = 0
+ 7 + 7   - 1 - 1
2x = 7 2x = -1
2 2 2 2
x = 3¹/₂     or     x = ⁻¹/₂

7 0

2 years ago

Read 2 more answers

Dana says that all even numbers are composite.<br><br> Is she correct?

Ksenya-84 [330]

Answer:

yes

Step-by-step explanation:

all even numbers greater than two are composite numbers

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=Expand%20%24%28a%20%2B%202%29%283a%5E2%20%2B%2012%29%28a%20-%202%29.%24" id="TexFormula1" titl

sashaice [31]

Answer:

3 $a^{4}$ - 48

Step-by-step explanation:

Given

(a + 2)(3a² + 12)(a - 2)

= (a + 2)(a - 2)(3a² + 12) ← expand the first pair of parenthesis using FOIL

=(a² - 4)(3a² + 12) ← expand using FOIL

= 3 $a^{4}$ + 12a² - 12a² - 48 ← collect like terms

= 3 $a^{4}$ - 48

5 0

3 years ago

If you roll a six-sided die, what is the probability that you will obtain a number greater than 2

Iteru [2.4K]

thee are 6 numbers on a die

4 are greater than 2

so you would have a 4/6, reduced to 2/3 probability

6 0

3 years ago