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2 years ago

8

Find the simplified product: 3/2x^5*3/64x^9

1 answer:

vlada-n [284]2 years ago

4 0

Answer: The answer is 4x^4 to the cube root of 2x^2

Brainliest plss!

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The circumference of a circle is 407 cm. Find the diameter, the radius, the length of an arc of 100°, and the area of a sector w

iogann1982 [59]

Answer: 407cm =

100° =

100° =

I hope this helps :)

4 0

3 years ago

A City's population rose 3% one year, 0.08 the next year, and by 2/50 the next year. Order these increases from least to greates

Mariulka [41]

3%, 2/50, 0.08 is ur answer!!

4 0

2 years ago

Read 2 more answers

P(x)equals=R(x)minus−C(x). Given R(x)equals=59 x minus 0.3 x squared59x−0.3x2 and Upper C left parenthesis x right parenthesi

Deffense [45]

Answer:

$P(x)=-0.3x^2+56x-14$

Step-by-step explanation:

The given functions are

$R(x)=59x-0.3x^2$

$C(x)=3x+14$

It is given that

$P(x)=R(x)-C(x)$

Substitute the values of given functions in the above equation.

$P(x)=59x-0.3x^2-(3x+14)$

$P(x)=59x-0.3x^2-3x-14$

Combine like terms.

$P(x)=-0.3x^2+(59x-3x)-14$

$P(x)=-0.3x^2+56x-14$

Therefore, the required function is $P(x)=-0.3x^2+56x-14$ .

4 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

A tennis ball travels at speed of 120 miles per hour.convert this rate to feet per second

Vanyuwa [196]

$120\frac{mile}{hour}\\\\
1 \ mile=5280\ feet\\
1\ hour=3600\ seconds\\\\
120\frac{mile}{hour}=120*\frac{5280}{3600}=\frac{120*5280}{3600}=\frac{633600}{3600}=176\frac{feet}{second}\\\\
\boxed{120\frac{mile}{hour}=176\frac{feet}{second}}$

4 0

3 years ago