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4 years ago

13

Consider 6x2 + 6x + 1. Which term immediately tells you that this expression is NOT a perfect square trinomial? Justify your ans

wer

1 answer:

Dennis_Churaev [7]4 years ago

5 0

Answer:

6x^2.

Step-by-step explanation:

The first term 6x^2, because 6 is not a perfect square.

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A rectangle, a triangle, and two congruent semicircles were used to form the figure shown. Rectangle 5cm,20cmcircle : radius 5cm

Schach [20]

Answer:

The area of a shape is the amount of space it occupies

Step 1:

We start by calculating the area of the rectangle using:

$\begin{gathered} A_{rectangle}=l\times b \\ \end{gathered}$

By substituting the values, we will have

$\begin{gathered} A_{rectangle}=l\times b \\ A_{rectangle}=20cm\times5cm \\ A_{rectangle}=100cm^2 \\ A_1=100cm^2 \end{gathered}$

Step 2:

we calculate the area of the triangle using:

$A_{triangle}=\frac{1}{2}\times base\times height$

By substituting the values, we will have

$\begin{gathered} A_{tr\imaginaryI angle}=\frac{1}{2}\times base\times he\imaginaryI ght \\ A_{tr\mathrm{i}angle}=\frac{1}{2}\times10cm\times8cm \\ A_{tr\mathrm{i}angle}=\frac{80cm^2}{2} \\ A_{tr\mathrm{i}angle}=40cm^2 \\ A_2=40cm^2 \end{gathered}$

Step 3:

Calculate the area of the two semicircles

$A_{semicircle}=\frac{\pi r^2}{2}$

By substituting the values, we will have

$\begin{gathered} A_{sem\imaginaryI c\imaginaryI rcle}=\frac{\pi r^{2}}{2} \\ A_{sem\mathrm{i}c\mathrm{i}rcle}=3.14\times\frac{5^2}{2} \\ A_{sem\mathrm{i}c\mathrm{i}rcle}=39.25cm^2 \\ Area\text{ of two semicircle will be} \\ A_3=39.25cm^2\times2 \\ A_3=78.5cm^2 \end{gathered}$

Step 4:

Calculate the area of the shape

We will calculate the area of the shape by adding all the individual areas together

$A_{shape}=A_1+A_2+A_3$

By substituting the values, we will have

$\begin{gathered} A_{shape}=A_{1}+A_{2}+A_{3} \\ A_{shape}=100cm^2+40cm^2+78.5cm^2 \\ A_{shape}=218.5cm^3 \end{gathered}$

Hence,

The area of the shape will be

$\Rightarrow218.5cm^2$

4 0

1 year ago

Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation?

Nady [450]

For this case we have the following polynomial:
$x ^ 2 = 12x - 15
$
The first thing to do is to place the variables on the same side of the equation.
We have then:
$x ^ 2 - 12x = -15
$
We complete the square by adding the term (b / 2) ^ 2 on both sides of the equation.
We have then:
$x ^ 2 - 12x + (-12/2) ^ 2 = -15 + (-12/2) ^ 2
$
Rewriting we have:
$x ^ 2 - 12x + (-6) ^ 2 = -15 + (-6) ^ 2

x ^ 2 - 12x + 36 = -15 + 36
 
(x-6) ^ 2 = 21$
$x-6 =+/- \sqrt{21}$
Therefore, the solutions are:
$x = 6 - \sqrt{21}$
$x = 6 + \sqrt{21}$
Answer:
the solution set of the equation is:
$x = 6 - \sqrt{21}$
$x = 6 + \sqrt{21}$

3 0

3 years ago

Read 2 more answers

How do I find the dimensions of one face

Sauron [17]

ok so dimensions is like area the dimensions of a cubes side is a one by one or the base and height of a cube is 1 inch each get it now or do you need me to explain more?

8 0

4 years ago

If a/b and x/y represent rational expressions and b0 and y0, what is true of their product? Select three options.

MAXImum [283]

Answer:

It’s the 2nd,3rd,and 4th

Step-by-step explanation:

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3 0

3 years ago

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HELP!!!<br> Write the slope and a point from each of the given equations<br><br> 1) y + 5 =-7/4(x-4)

blagie [28]

Answer: x=

7

Step-by-step explanation:

7 0

3 years ago