Ask question

Ask question

All categories

3 years ago

15

Factor 64xº - 27

1 answer:

kumpel [21]3 years ago

4 0

Answer:

Step-by-step explanation:

a³-b³=(a-b)(a²+ab+b²)

64x³-27=(4x)³-3³=(4x-3)((4x)²+4x×3+3²)=(4x-3)(16x²+12x+9)

You might be interested in

The pirouette dance team needs more than $300 to cover

noname [10]

Answer:

You need "more than" 300.

Step-by-step explanation:

i did the test

8 0

2 years ago

Determine the correct scientific notation form of the number.

Dahasolnce [82]

0.000000084

we need to move the decimal 8 places to the right

8.4* 10^-8

3 0

3 years ago

Read 2 more answers

A rectangular prism and a square pyramid were joined to form a composite figure.

azamat

Answer:

B. 333 in.2

Step-by-step explanation:

The area of the base is 81^2

lateral is 45 x 4 = 180^2 (9x5x4)

180^2 add the 72 pyramid = 252^2 + base of 81^2 = 333^2

Composite figueres i do like this as shown to you a few seconds ago.

The triangle shows us just the height

4 inches

We can see that height is smaller central isosceles height across the center base point.

We also can remember to use the length 9inches but divide by 2 and get each triangle area this way.

4 x 1/2 base = 4x 1/2 4.5 = 4 x 2.25 = 9^2 each right side triangle

9 x 8 = 72^2

we add the areas 72+ 81+lateral 180 = 333 inches^2

8 0

3 years ago

I told you people that i would be sending out a series of problems todayy!!

qwelly [4]

Each student would need to raise $27.56

3 0

3 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

3 years ago