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user100 [1]
3 years ago
12

Sorry I forgot to add more points please do all and hurry

Mathematics
1 answer:
denis-greek [22]3 years ago
3 0
To find the area of a quadrilateral, multiply the length by the width.
You might be interested in
How do you solve the following equation: 5-(2x-3)=-8+2x
kirza4 [7]
I believe it would be done like this:
subtract 2x-8 from both sides. then use cubric formula.
answer should be  x= 1.688242

4 0
3 years ago
There is a group of student,1/3 are in elementary ,1/2 are in middle school and 15 are in high school. How many are in the group
OverLord2011 [107]
1/2 + 1/3 = 3/6 + 2/6 = 5/6
we have 5/6 so 15 is 1/6 of the entire group
15 × 6 = 90
7 0
3 years ago
Which term best describes lines that meet at right angles ?
Sholpan [36]

Answer:

Perpendicular

Step-by-step explanation:

Perpendicular lines meet at a 90 degree angle.

5 0
3 years ago
Not sure what to do here can someone please help
kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

  \sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

  u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}

Solutions to this equation are ...

  u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

  x = u² -2 = 2² -2

  x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0
1 year ago
Could a polyhedron exist with the given number of faces, vertices, and edges? Drag yes or no to each combination
DanielleElmas [232]

Answer:

(a) No

(b) No

(c) No

Step-by-step explanation:

Given

See attachment

Required

Select Yes or No for each

To do this, we make use of Euler's formula

F + V - E = 2

Where

F \to Faces; V \to Vertices; E \to Edges

(a):\ Faces = 8; Vertices = 12; Edges = 6

Using: F + V - E = 2

8 + 12 - 6 = 2

14 = 2

<em>The above equality is false. Hence, (a) does not exist</em>

(b):\ Faces = 6; Vertices = 6; Edges = 4

Using: F + V - E = 2

6 + 6 - 4 = 2

8 = 2

<em>The above equality is false. Hence, (b) does not exist</em>

<em />

<em />(c):\ Faces = 20; Vertices = 30; Edges = 12<em />

Using: F + V - E = 2

20 + 30 - 12 = 2

38 = 2

<em>The above equality is false. Hence, (c) does not exist</em>

5 0
2 years ago
Read 2 more answers
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