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GuDViN [60]
2 years ago
7

Use the exponent properties to simplify the following

z%5E3%7D" id="TexFormula1" title="\frac{3x^3y^2z}{27xy^2z^3}" alt="\frac{3x^3y^2z}{27xy^2z^3}" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
7nadin3 [17]2 years ago
7 0

Step-by-step explanation:

x^2/9 z^2

hiiii checkout the steps

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Javier bought a new tv for $479.76. he will make equal payments each month for 2 years.how can javier use compatible numbers to
Fiesta28 [93]
12 months 2 years = 24 months. so 479.76÷24 (because of how much gies into each month) = 19.99 each month. so round it to 20$ each month
3 0
3 years ago
Help OvO with this math question-
harkovskaia [24]

Answer:

C

Step-by-step explanation:

40 x 80 = 3200

120 x 100 = 12000

3200 + 12000 = 15200

6 0
2 years ago
Read 2 more answers
What is the range of the function f(x)=7-3x when the domain is {-4, -2, 0, 2}?
klemol [59]

Answer:

{19,13,7,1}

Step-by-step explanation:

Since the domain is the input we have to plug in each of those numbers to get the range ( or the output.) When we plug in -4 we get 19. When we plug in -2 we get 13, when we plug in 0 we get 7 and when we plug in 2 we get 1. (:

6 0
2 years ago
2. The table below shows that the distance d varies directly as the time t. Find the constant of variation and the equation whic
Artemon [7]
<h2><u>Problem Solving</u>:-</h2>

2. The table below shows that the distance d varies directly as the time t. Find the constant of variation and the equation which describes the relation.

<h2><u>Solution</u>:-</h2>

Since the distance d varies directly as the time t, then d = kt.

Using one of the pairs of values, (2, 20), from the table, substitute the values of d and t in d = kt and solve for k.

\sf{\rightarrow{d =  kt}}

\sf\rightarrow{20  = 2k }

\sf\rightarrow{K= \frac{20}{2} }

\sf\rightarrow{K={\color{magenta}{10}}}

<h2><u>Answer</u>:-</h2>
  • Therefore, the constant of variation is 10.
3 0
2 years ago
Suppose a white dwarf star has a diameter of approximately 1.8083 to the power of 4 km. Use the formula 4n to the power of 2 to
aivan3 [116]

ANSWER:

The surface area of the star is 3.2700 x 10^{8} square kilometres.

EXPLANATIONS:

Diameter of the star = 1.8083 x 10^{4} Km.

Surface area of the star = 4n^{2}

Where n is the radius of the star.

So that;

n = \frac{ 1.8083 * 10^{4} }{2}

   = 0.90415 x 10^{4}

n =  0.90415 x 10^{4} Km

Thus,

Surface area = 4 x (0.90415*10^{4} )^{2}

                     = 326994889

Surface area = 3.2700 x 10^{8} km^{2}

Therefore, the surface area of the star is 3.2700 x 10^{8} square kilometres.

4 0
2 years ago
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