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3 years ago

Simplificar expresiones lineales-2+7m-7m+5m

Mathematics

1 answer:

Helen [10]3 years ago

8 0

Answer:

I am not sure it correct or not but I try to answer

Step-by-step explanation:

0-2+7m-7m+5m=7

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What is the surface area?

notsponge [240]

Answer:

369 m²

Step-by-step explanation:

Area of Triangle (sides): 1/2bh

=1/2(bh)

=1/2 (9•16)

=1/2 (144)

=72

There are four sides of the prism:

72•4=288

Area of Square (base): l•w

=9•9

=81

Surface Area = 288+81=369 m²

5 0

3 years ago

Help me out guys. Got stuck over this algebra question

vovikov84 [41]

Answer:

Compare LHS and RHS to prove the statement.

Step-by-step explanation:

Given: a + b + c = 0

We have to show that $\frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} + \frac{1}{1 + x^a + x^{-b}} = 1$

We take LCM, simplify the terms and compare LHS and RHS. We will see that LHS = RHS and the statement will be proved.

Taking LCM, we get:

$\frac{(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1)}{(1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})}$ = 1

⇒ $(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1) = (1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})$

We simplify each term and then compare LHS and RHS.

Simplifying the first term:

$(1 + x^c + x^{-a})(1 + x^a + x^{-b})$

= $x^{c + a} + x^{c - b} + x^c + 1 + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 1$

= $x^{c + a} + x^{c - b} + x^{c} + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 2 \hspace{5mm} \hdots (A)$

Now, we simplify the second term we have:

$(1 + x^b + x^{-c})(x^a + x^{-b} + 1)$

= $x^{a + b} + 1 + x^{b} + x^{a - c} + x^{-b - c} + x^{-c} + x^{a} + x^{-b} + 1$

= $x^{a + b} + x^{b} + x^{a - c} + x^{- c - b} + x^{-c} + x^a + x^{-b} + 2 \hspace{5mm} \hdots (B)$

Again, simplifying $(x^b + x^{-c} + 1)(x^c + x^{-a} + 1)$ ,

= $x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1$

= $x^{b + c} + x^{b - a} + x^b + x^{-c -a} + x^{-c} + x^c + x^{-a} + 2$ (C)

Therefore, LHS = A + B + C

= $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

Similarly. RHS

= $(x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1)(x^a + x^{-b} + 1)$

Note that if a + b + c = 0, $\implies x^{a + b + c} = x^0 = 1$

So, RHS = $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

We see that LHS = RHS.

Therefore, the statement is proved.

3 0

4 years ago

A two-variable inequality is shown in the graph.

sammy [17]

The point which is not included in the solution set for the inequality of the two-variable inequality shown in the graph is (-1,3).

<h3>How to graph the inequality?</h3>

Inequality of a graph is represented with the greater then(<), less then(>) or with the other inequity signs. The inequality line on the graph is represented with the dotted lines.

A two-variable inequality is shown in the graph. It is a parabola. The vertex form of parabola is the equation form of quadratic equation which is used to find the coordinate of vertex points at which the parabola crosses its symmetry.

The standard equation of the vertex form of parabola is given as,

$y=a(x-h)^2+k$

Here, (h, k) is the vertex point. In the graph, vertex points are (1,2). Thus, the equation become,

$y=a(x-1)^2+2\\$

By the point of graph (0,3), find the value of a,

$3=a(0-1)^2+2\\3=a+2\\a=3-2\\a=1$

Thus, the equation become,

$y=1(x-1)^2+2\\y=x^2-2x+1+2\\y=x^2-2x+3\\$

The graph of this line is attached below. In this graph only point (-1,3), does not fall hence, does not satisfy the equation.

Thus, the point which is not included in the solution set for the inequality of the two-variable inequality shown in the graph is (-1,3).

Learn more about the graph of the inequality here;

brainly.com/question/62792

#SPJ1

5 0

2 years ago

A -5/4 B -8/10 C 3/4 D -5/3

Ede4ka [16]

Well let's remember csc = hypotenuse/opposite

Let's use the distance formula to find the length of the hypotenuse.
√(0−−8)^2+(−6−0)^2
√(0+8)^2+(−6+0)^2
√(8)^2+(−6)^2
√64+36
√100
H = 10

so csc = 10/opposite

Now to find the opposite length using distance formula
√(0−−8)^2+(0−0)^2
√(0+8)^2+(0+0)^2
√(8)^2+(0)^2
√64+0
√64
O = 8

So csc = 10/8 or 1.25

3 0

3 years ago

If 4 1/2 = 2, 8 1/1= 2, and 16 then for what value of xwould x 1/5 = 2? 24 32 02 04

nika2105 [10]

Answer:

D) 0.0004

Step-by-step explanation:

(0.02)^2=0.02*0.02=0.0004

7 0

3 years ago

Simplificar expresiones lineales-2+7m-7m+5m​

Simplificar expresiones lineales-2+7m-7m+5m