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2 years ago

3x^3 - 2y^-6x^2y^2+xy for x=2/3 and y=1/2

Mathematics

1 answer:

Dominik [7]2 years ago

4 0

Answer:

-13

Step-by-step explanation:

3x^3 - 2y^-6x^2y^2+xy

x=2/3 and y=1/2

Solve :

3x^3 - 2y^-6x^2y^2+xy

Put value of x and y in this equation

= 3(2/3)^3 - 2(1/2)^-6(2/3)^2(1/2)^2+(2/3)(1/2)

= -13

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Jordan received 9 text messages last week.She received 3 times more text message this week than last week. How many text message

vazorg [7]

If she got 9 messages last week and 3rimes as much this week then she got 27 messages because 9x3=27

8 0

3 years ago

Read 2 more answers

Really need helpppp?!

user100 [1]

Answer:

6.43

pls follow me and mark brainliest ...thank u

Step-by-step explanation:

length of side x will be

Tan tita = opp/adjacent

tan65 = x/3

tan 65 is 2.14

so...you are left with 2.14=x/3

by cross multiplying

x= 2.14 × 3

= 6.43

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2 years ago

HELP NOW!!!!

VMariaS [17]

Using the concept of probability and the arrangements formula, there is a

0.002 = 0.2% probability that the first 8 people in line are teachers.

----------------------------------

A probability is the number of desired outcomes divided by the number of total outcomes.
The order in which they are positioned is important, and all people will be positioned, and thus, the arrangements formula is used to find the number of outcomes.

The number of possible arrangements from a set of n elements is given by:

$A_n = n!$

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The desired outcomes are:

First 8 people are teachers, in 8! possible ways.
Last 4 are students, in 4! possible ways.

Thus, $D = 8! \times 4!$

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For the total outcomes, number of arrangements of 12 people, thus:

$T = 12!$

The probability is:

$p = \frac{D}{T} = \frac{8! \times 4!}{12!} = 0.002$

0.002 = 0.2% probability that the first 8 people in line are teachers.

A similar problem is given at brainly.com/question/24650047

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3 years ago

ABC is an isosceles triangle: AC = 13cm. What is the length of AB?

sveticcg [70]

AB=AC

AC=13cm

$\\ \sf\longmapsto AB$

$\\ \sf\longmapsto AC$

$\\ \sf\longmapsto 13cm$

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2 years ago

Select all the correct equations.

Naya [18.7K]

Answer:

You did not post the options, but i will try to answer this in a general way.

Because we have two solutions, i know that we are talking about quadratic equations, of the form of:

0 = a*x^2 + b*x + c.

There are two easy ways to see if the solutions of this equation are real or not.

1) look at the graph, if the graph touches the x-axis, then we have real solutions (if the graph does not touch the x-axis, we have complex solutions).

2) look at the determinant.

The determinant of a quadratic equation is:

D = b^2 - 4*a*c.

if D > 0, we have two real solutions.

if D = 0, we have one real solution (or two real solutions that are equal)

if D < 0, we have two complex solutions.

3 0

3 years ago