10. APPLICATION Inspector 47 at the Zap battery plant

The domain of a relation R is the set of real numbers. x is related to y under relation R if dxe ≤ dye . Select the description

Juli2301 [7.4K]

Answer:

Step-by-step explanation:u

8 0

2 years ago

A golden rectangle is 32cm long.The ratio of length to width is (1 + square root of 5) :2. What is the width of the rectangle in

zhannawk [14.2K]

Answer:

Width = 16(√5) - 1

Step-by-step explanation:

We are told that the golden rectangle is 32 cm long.

Thus, length = 32 cm

We are also told that the ratio of the length to the width is; (1 + √5):2

Thus;

If a length of (1 + √5) yields a width of 2

Then, a length of 32 cm would yield a width of; (32 x 2)/(1 + √5)

So corresponding width = 64/(1 + √5)

We want to reduce this width to it's simplest radical form which means the denominator should have no square root.

Thus, let's multiply top and bottom by (1 - √5);

Width = 64 x (1 - √5)/[(1 + √5) x (1 - √5)]

Width = 64(1 - √5)/(1 - 5)

Width = 64(1 - √5)/(-4)

Width = -16(1 - √5)

Width = 16(√5 - 1)

Width = 16√5 - 1

8 0

2 years ago

The square of a number decreased by 3 times the number 28 find all possible values for the number

stealth61 [152]

Question:

The square of a number decreased by 3 times the number is 28 find all possible values for the number

Answer:

The possible values of number are 7 and -4

Solution:

Given that the square of a number decreased by 3 times the number is 28

To find: all possible values of number

Let "a" be the unknown number

From given information,

square of a number decreased by 3 times the number = 28

$a^2 - 3a = 28$

$a^2 - 3a - 28 = 0$

Let us solve the above quadratic equation

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text { For } a^{2}-3 a-28=0 \text { we have } a=1, b=-3, c=-28$

$\begin{aligned}&a=\frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-28)}}{2 \times 1}\\\\&a=\frac{3 \pm \sqrt{9+112}}{2}\\\\&a=\frac{3 \pm \sqrt{121}}{2}=\frac{3 \pm 11}{2}\\\\&a=\frac{3+11}{2} \text { or } a=\frac{3-11}{2}\\\\&a=7 \text { or } a=-4\end{aligned}$

Thus the possible values of number are 7 and -4

5 0

2 years ago

Diane reed had office supply sales of $3,285 this month. this is a 78% of last months supply sales. what was the amount of last

Hoochie [10]

To solve this problem, we can set up an equation, letting the unknown value of last month's supply sales be represented by the variable x.

First, we must convert 78% to its decimal equivalent, by dividing 78/100. This is because percentages are parts of a whole 100 percent.

78/100 = 0.78

Now, we are ready to set up our equation, using our knowledge that the word "of" refers to multiplication in mathematics.

3,285 = 0.78x

To solve our equation, we must divide both sides of the equation by 0.78 to get the variable x alone on the right side of the equation.

x = 4211.53846154

This means that last month's supply sales were about $4,211.54 (we round the last digit up because 8 is greater than 5).

Hope this helps!

8 0

2 years ago

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2

Margaret [11]

<u>Answer-</u>

$\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}$

<u>Solution-</u>

Given expression is $(x+2)^4$

Applying Binomial Theorem

$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here,

a = x, b = 2 and n = 4

So,

$\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i$

Expanding the summation

$=\dfrac{4!}{0!\left(4-0\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(4-1\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(4-2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(4-3\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(4-4\right)!}x^0\cdot \:2^4$

$=\dfrac{4!}{0!\left(4\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(3\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(1\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(0\right)!}x^0\cdot \:2^4$

$=1\cdot x^4\cdot \:1+4\cdot x^3\cdot \:2+6x^2\cdot \:4+4\cdot x\cdot \:8+1\cdot 1\cdot \:16$

$=x^4+8x^3+24x^2+32x+16$

4 0

2 years ago

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