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3 years ago

What is the relationship between the value of the digit 3 in 4,231 and in the value of the digit 3 in the number 3,421?

Mathematics

2 answers:

MrRa [10]3 years ago

7 0

Answer:

In 4231 the value of digit 3 is 1/100 the value of digit 3 in 3421..

hope this helped you

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leva [86]3 years ago

5 0

Answer:

In 4,231, the value of the digit 3 is 100 times the value of the digit 3 in 3,421.

Step-by-step explanation:

Hope this helps.

Read more about expressions at:

brainly.com/question/723406

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8 0

2 years ago

Jared has 4 less than 3 times the number of trophies that zach had.

dimulka [17.4K]

How much trophies does Jared have(

4 0

3 years ago

Read 2 more answers

Jane earns $7.50 an hour and time-and-a-half for overtime. How

ArbitrLikvidat [17]

Answer:

339.375 choose d

Step-by-step explanation:

Regular work 40 hours and 3.5 overtime hours

Overtime wage is 1.5 times the regular one

40×7.5+3.5×7.5×1.5=339.375

4 0

3 years ago

Determine consecutive integer values of x between which each real zero is located.

frozen [14]

Answer:

1. x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

2. x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

Step-by-step explanation:

Solve for x:

x^3 + 6 x^2 + 6 x - 4 = 0

The left hand side factors into a product with two terms:

(x + 2) (x^2 + 4 x - 2) = 0

Split into two equations:

x + 2 = 0 or x^2 + 4 x - 2 = 0

Subtract 2 from both sides:

x = -2 or x^2 + 4 x - 2 = 0

Add 2 to both sides:

x = -2 or x^2 + 4 x = 2

Add 4 to both sides:

x = -2 or x^2 + 4 x + 4 = 6

Write the left hand side as a square:

x = -2 or (x + 2)^2 = 6

Take the square root of both sides:

x = -2 or x + 2 = sqrt(6) or x + 2 = -sqrt(6)

Subtract 2 from both sides:

x = -2 or x = sqrt(6) - 2 or x + 2 = -sqrt(6)

Subtract 2 from both sides:

Answer: x = -2 or x = sqrt(6) - 2 or x = -2 - sqrt(6)

_________________________________________

Solve for x:

x^4 - 2 x^3 - 6 x^2 + 8 x + 5 = 0

Eliminate the cubic term by substituting y = x - 1/2:

5 + 8 (y + 1/2) - 6 (y + 1/2)^2 - 2 (y + 1/2)^3 + (y + 1/2)^4 = 0

Expand out terms of the left hand side:

y^4 - (15 y^2)/2 + y + 117/16 = 0

Subtract -3/2 sqrt(13) y^2 - (15 y^2)/2 + y from both sides:

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

y^4 + (3 sqrt(13) y^2)/2 + 117/16 = (y^2 + (3 sqrt(13))/4)^2:

(y^2 + (3 sqrt(13))/4)^2 = (3 sqrt(13) y^2)/2 + (15 y^2)/2 - y

Add 2 (y^2 + (3 sqrt(13))/4) λ + λ^2 to both sides:

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

(y^2 + (3 sqrt(13))/4)^2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (y^2 + (3 sqrt(13))/4 + λ)^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = -y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2

-y + (3 sqrt(13) y^2)/2 + (15 y^2)/2 + 2 λ (y^2 + (3 sqrt(13))/4) + λ^2 = (2 λ + 15/2 + (3 sqrt(13))/2) y^2 - y + (3 sqrt(13) λ)/2 + λ^2:

(y^2 + (3 sqrt(13))/4 + λ)^2 = y^2 (2 λ + 15/2 + (3 sqrt(13))/2) - y + (3 sqrt(13) λ)/2 + λ^2

Complete the square on the right hand side:

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2 + (4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1)/(4 (2 λ + 15/2 + (3 sqrt(13))/2))

To express the right hand side as a square, find a value of λ such that the last term is 0.

This means 4 (2 λ + 15/2 + (3 sqrt(13))/2) (λ^2 + (3 sqrt(13) λ)/2) - 1 = 8 λ^3 + 18 sqrt(13) λ^2 + 30 λ^2 + 45 sqrt(13) λ + 117 λ - 1 = 0.

Thus the root λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) allows the right hand side to be expressed as a square.

(This value will be substituted later):

(y^2 + (3 sqrt(13))/4 + λ)^2 = (y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)))^2

Take the square root of both sides:

y^2 + (3 sqrt(13))/4 + λ = y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) - 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2)) or y^2 + (3 sqrt(13))/4 + λ = -y sqrt(2 λ + 15/2 + (3 sqrt(13))/2) + 1/(2 sqrt(2 λ + 15/2 + (3 sqrt(13))/2))

Solve using the quadratic formula:

y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) + sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 - 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) or y = 1/4 (sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13))) - sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13))) or y = 1/4 (-sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)) - sqrt(2) sqrt((108 - 24 sqrt(13) λ - 16 λ^2 + 4 sqrt(2) sqrt(4 λ + 15 + 3 sqrt(13)))/(4 λ + 15 + 3 sqrt(13)))) where λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3))

Substitute λ = 1/4 (-3 sqrt(13) - 5) + (2 2^(2/3) (i sqrt(3) + 1))/(i sqrt(183) - 29)^(1/3) + ((-i sqrt(3) + 1) (i sqrt(183) - 29)^(1/3))/(2 2^(2/3)) and approximate:

y = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x - 1/2 = -2.60947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or y = -0.984343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x - 1/2 = -0.984343 or y = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or y = 1.17884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x - 1/2 = 1.17884 or y = 2.41497

Add 1/2 to both sides:

x = -2.10947 or x = -0.484343 or x = 1.67884 or y = 2.41497

Substitute back for y = x - 1/2:

x = -2.10947 or x = -0.484343 or x = 1.67884 or x - 1/2 = 2.41497

Add 1/2 to both sides:

Answer: x = -2.10947 or x = -0.484343 or x = 1.67884 or x = 2.91497

8 0

3 years ago

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

5 0

3 years ago