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Ira Lisetskai [31]
3 years ago
13

Factor the expression. d2 − 4d + 4

Mathematics
2 answers:
sesenic [268]3 years ago
7 0

Answer:

(d - 2)²

Step-by-step explanation:

d² - 4d + 4 ← is a perfect square of the form

(x ± a )²  = x² ± 2ax + a²

compare coefficients of like terms

- 2a = - 4 ⇒ a =  2 and a² = 4, hence

d² - 4d + 4 = (d - 2)²


ankoles [38]3 years ago
6 0

Rewrite it in the form a^2 - 2ab + b^2, where a = d and b = 2

d^2 - 2(d)(2) + 2^2

Use the Square of Difference: (a - b)^2 = a^2 - 2ab + b^2

(d - 2)^2

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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
Question 9(Multiple Choice Worth 4 points)
Dmitrij [34]

Answer:

The graph will shift 9 units to the right

Sorry about my hand writing

8 0
3 years ago
A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na
ira [324]

Answer:

Approximately 58.2\; \text{nautical miles} (assuming that the bearing is {\rm S$29^{\circ}$W}.)

Step-by-step explanation:

Let v denote the speed of the ship, and let t denote the duration of the trip. The magnitude of the displacement of this ship would be v\, t.

Refer to the diagram attached. The direction {\rm S$29^{\circ}$W} means 29^{\circ} west of south. Thus, start with the south direction and turn towards west (clockwise) by 29^{\circ} to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of v\, t. The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly 29^{\circ}.

Since the hypotenuse is of length v\, t, the dashed line segment opposite to the \theta = 29^{\circ} vertex would have a length of:

\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}.

Substitute in \begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned} and t = 6\; \text{hour}:

\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}.

7 0
2 years ago
The population of a town gradually decreases by about 30
topjm [15]
It would be 3,600 - 30 = 3570
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2 years ago
A telephone company offers a monthly cellular phone plan for $ 39.99 It includes 300 anytime minutes plus $ 0.25 per minute for
Nataly [62]
A. 39.99 
b. 39.99 + 0.25(35) = 48.74
c. 39.99 + 0.25(1) = 40.24
3 0
3 years ago
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