The side of an equilateral triangle is 12. Its area is
Select one:
a. 144
b. 72
c. 36 √3
d. 48

Answers

Answer 1

Answer:

area is 1/2bh

in an euqilateral triangle, all sides are equal so base and height are 12

1/2 x 12 x 12

1/2 x 144= 72 (B)

Answer 2

Answer: Choice C. [tex]36\sqrt{3}[/tex]

This is the same as writing 36*sqrt(3)

===================================================

Work Shown:

x = 12 = side length of equilateral triangle

A = area of equilateral triangle

A = 0.25*sqrt(3)*x^2

A = 0.25*sqrt(3)*12^2

A = 0.25*sqrt(3)*144

A = 0.25*144*sqrt(3)

A = 36*sqrt(3) .... answer is choice C

Side note: the area approximates to 62.354 square units.


Related Questions


HELP PLEASE IM STUCK!

1. Which of these describes the relation for this set of coordinate pairs?
{(-1, 5), (12, 18), (0, 6), (-3, 3), (4, ?), (?, 11)}
a. x - y = 6 b. f(x) = x +6 c. f(x) = 6 d. y = 6x e. None of these

Answers

Answer:

b) f(x) = x + 6

Step-by-step explanation:

The coordinate (0, 6) makes the y-intercept = 6. Only one of these functions has that intercept: f(x) = x + 6. If you plug in each coordinate the outputted y-value matches up, making this the right answer.

HELP MATH ANSWER QUESTION

Answers

Answer:

x = 84 , y = 168

Step-by-step explanation:

[tex]Sin 60 = \frac{opposite side}{hypotenuse}\\\\ \frac{\sqrt{3}}{2}= \frac{AB}{BC}\\\\ \frac{\sqrt{3}}{2} = \frac{84\sqrt{3}}{y}\\\\Cross\ multiply ,\\\\y*\sqrt{3} =2*84 \sqrt{3}\\\\y= \frac{2*84*\sqrt{3}}{\sqrt{3}}\\\\y=168[/tex]

[tex]Cos\60 = \frac{adjacent}{hypotenuse}\\\\ \frac{1}{2}= \frac{AC}{BC}\\\\ \frac{1}{2}= \frac{x}{168}\\\\ \frac{1}{2}*168=x\\\\x= 84[/tex]

Which relationship is always true for the angles x,y and z of triangle ABC

Answers

Answer:

B. y + z = x

Step-by-step explanation:

x is an exterior angle of the triangle.

y and z are the opposite angles opposite the exterior angle.

The exterior angle theorem of a triangle states that the measure of an exterior angle equals the measure of the sum of the two angles opposite the exterior angle.

Thus:

y + z = x

An expression is shown below:

3(m + 5 + 9m)

Part A: Write two expressions that are equivalent to the given expression. (3 points)

Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties. Explain which properties you used. (4 points)

Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m. (3 points)

Answers

The answers are as follows part A = 30m+15  part B =3m+15+27m

and partC = 30m+15

What is an expression?

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition,substraction, multiplication and division.

Part A:- Two expressions that are equivalent to the given expressions are:-

3m   +   15   +   27m

30m  +   15

Part B:  Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties.

3  (  m   +   5    +   9m   )

Open the bracket by multiplying 3 by what is in the bracket

3m   +   15   +   27m

Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m.

3   (  m    +   5   +   9m  )

Open the bracket by multiplying 3 by what is in the bracket

3m    +    15    +     27m

Collect like terms together

3m    +     27m  +   15

=   30m   +    15

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A box with a square base and open top must have a volume of 256000 c m 3 . We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only x , the length of one side of the square base.

Answers

Answer:

Follows are the response to the given question:

Step-by-step explanation:

The volume of the box:

[tex]V = x\times x \times h = 256000 \ cm^3\\\\\to x^2 \times h = 256000\\\\\to h = \frac{256000}{x^2}[/tex]  

The surface area of the open box is:

[tex]A(x) = x \times x + 2 \times (x \times h +x \times h)\\\\A(x) = x^2 + 4 \times x \times h\\\\A(x) = x^2 + \frac{1024000}{x}\\\\\frac{d(x^n)}{dx} = n \times x^{(n - 1)}\\\\[/tex]

Use above formula

[tex]A'(x) = 2 \times x - \frac{1024000}{x^2}\\\\[/tex]

[tex]A'(x) = 0\\\\2\times x - \frac{1024000}{x^2} = 0\\\\2x = \frac{1024000}{x^2}\\\\x^3 = 512000\\\\x = (512000)^{(\frac{1}{3})} = 80\ cm\\\\[/tex]

Now

[tex]A''(x) = 2\times 1 + 2\times \frac{1024000}{x^3}\\\\A''(x) = 2 + \frac{2048000}{x^3}\\\\x = 80 \ cm\\\\A''(80) = 2 + \frac{2048000}{80^3} = 6\\\\[/tex]

therefore [tex]A"(x) > 0,[/tex] x amount of material used in minimum.

[tex]h = \frac{256000}{80^2} = 40\ cm[/tex]

CAN SOMEONE PLEASE HELP ME GOOD, I need this to graduate ): 5. Given that AABC - ADEC, find the
value of x.

Answers

Answer:

ans: 4

Step-by-step explanation:

corresponding sides are proportional since given triangle are similar triangle, I.e

(4/5.5) = { (2x+8)/(6x-2)}

8/11 = ( x+ 4 ) / ( 3x - 1 )

8( 3x - 1 )= 11( x + 4 )

24x - 8 = 11x + 44

13x = 52

x = 4

Select the correct answer.
At a high school there are 53 players on the football team, 15 players on the baseball team, and 12 players on
the basketball team. How many ways can a committee be formed with 1 representative from each team?
Math

Answers

Answer:

53*15*12=9540

Step-by-step explanation:

its just 53 times 15 times 12 for all the possibilities.

Say the first football player was picked, same with the first baseball player, and the first basketball player were all picked, then another possiblity would be the first football player, the second baseball player, and the first basketball player, here is a numerical example.

Football            Baseball             Basketball

     1                         1                             1  

     1                         2                            1

     1                         2                            2

     1                         2                            3

     1                         2                            4

and so on including all the patterns it would be 9540 possibilities

The total number of ways of forming a committee by selecting one representative from each team is 9540.

What is combination?

A combination is a mathematical technique that determines the number of possible arrangements or the number of ways  in a collection of items where the order of the selection does not matter.

Combination Formula

[tex]nC_{r}= \frac{n!}{r!(n-r)!}[/tex]

Where,

[tex]nC_{r}[/tex] is a number of combination.

n is total number of objects in the set.

r is the number of choosing objects from the set.

Multiplication rule in combination?

According to the multiplication rule in combination if there are a ways of doing something and b ways of doing another thing, then there are a · b ways of performing both actions.

According to the given question

Total number of football players = 53

Total number of baseball players = 15

Total number of basket ball players = 12

Therefore,

Number of ways of selecting one representative from football team is given by  

[tex]53C_{1} =\frac{53!}{1!52!} = 53[/tex]

Number of ways of selecting one representative from baseball team is given by

[tex]15C_{1} =\frac{15!}{1!14!}=15[/tex]

Number of ways of selecting one representative from basketball team is given by

[tex]12C_{1} =\frac{12!}{1!11!} =12[/tex]

So, the total number of ways of forming a committee by selecting one representative from each team = 53 × 15 × 12 =9540.

Hence,  total number of ways of forming a committee by selecting one representative from each team 9540.

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What is the probability of rolling a number less than or equal to 8 with the
sum of two dice, given that at least one of the dice must show a 6?

Answers

Answer:

I hope this helps

the outcomes are the compulsory 6, and 1 or 2

Step-by-step explanation:

[tex] \frac{3}{6} \\ \frac{1}{2} or \: 0.5[/tex]

Segment [tex]$s_1$[/tex] has endpoints at [tex]$(3+\sqrt{2},5)$[/tex] and[tex]$(4,7)$[/tex]. Segment [tex]$s_2$[/tex] has endpoints at [tex]$(6-\sqrt{2},3)$[/tex] and[tex]$(3,5)$[/tex]. Find the midpoint of the segment with endpoints at the midpoints of [tex]$s_1$[/tex] and [tex]$s_2$[/tex]. Express your answer as [tex]$(a,b)$[/tex].

Answers

Answer:

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).

Step-by-step explanation:

Midpoint of a segment:

The coordinates of the midpoint of a segment are the mean of the coordinates of the endpoints of the segment.

Midpoint of s1:

Using the endpoints given in the exercise.

[tex]x = \frac{3 + \sqrt{2} + 4}{2} = \frac{7 + \sqrt{2}}{2}[/tex]

[tex]y = \frac{5 + 7}{2} = \frac{12}{2} = 6[/tex]

Thus:

[tex]M_{s1} = (\frac{7 + \sqrt{2}}{2},6)[/tex]

Midpoint of s2:

[tex]x = \frac{6 - \sqrt{2} + 3}{2} = \frac{9 - \sqrt{2}}{2}[/tex]

[tex]y = \frac{3 + 5}{2} = \frac{8}{2} = 4[/tex]

Thus:

[tex]M_{s2} = (\frac{9 - \sqrt{2}}{2}, 4)[/tex]

Find the midpoint of the segment with endpoints at the midpoints of s1 and s2.

Now the midpoint of the segment with endpoints [tex]M_{s1}[/tex] and [tex]M_{s2}[/tex]. So

[tex]x = \frac{\frac{7 + \sqrt{2}}{2} + \frac{9 - \sqrt{2}}{2}}{2} = \frac{16}{4} = 4[/tex]

[tex]y = \frac{6 + 4}{2} = \frac{10}{2} = 5[/tex]

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).

At Dubai English School, 549 students use buses to go to school. If this number is 75% of the total school enrollment, then how many students are enrolled in total?

Answers

732 is the answer have a nice day and I hope you pass your exam or complete your homework

Instructions: Find the value of x
I’ll mark brainliest please help

Answers

The little lines theu each section are telling you all those sections are identical, which mean they are the same length.

You are told one section is 10, which means x is also 10

X = 10

Are these triangles congruent?

Answers

Answer:

yes...

Step-by-step explanation:

its a congrate triangle

Yes the triangles are congruent because they have the same sides but not the same lengths and width.

A researcher has obtained the number of hours worked per week during the summer for a sample of 15 students. 40 25 35 30 20 40 30 20 40 10 30 20 10 5 20 Using this data set, compute the following: a. Median b. Mean c. Mode d. 40th percentile e. Range f. Sample variance g. Standard deviation

Answers

Answer:

Mean = 25

Median = 25

Mode = 6

Step-by-step explanation:

Given the data :

Using calculator :

Mode = highest occurring data point

Range = max - min = 40 - 5 = 35

Sample variance = 128.57 (calculator)

Sample standard deviation = sqrt(variance) = 11.34

Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights of oranges are also normally distributed with a mean of 131 grams and a standard deviation of 20 grams. Amy has an apple that weighs 90 grams and an orange that weighs 155 grams.

Required:
a. Find the probability a randomly chosen apple exceeds 100 g in weight.
b. What weight do 80% of the apples exceed?

Answers

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that [tex]\mu = 85, \sigma = 8[/tex]

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{100 - 85}{8}[/tex]

[tex]Z = 1.875[/tex]

[tex]Z = 1.875[/tex] has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.84 = \frac{X- 85}{8}[/tex]

[tex]X - 85 = -0.84*8[/tex]

[tex]X = 78.28[/tex]

The weight that 80% of the apples exceed is of 78.28g.

Find Length of x line OR

Answers

ATQ:-

Both figures r similar (given )

so :-

[tex] \frac{5}{2.5} = \frac{3}{1.5} = \frac{2}{1} = \frac{4}{x} \\ \frac{2}{1} = \frac{4}{x} \\ \frac{ \cancel{2}}{1} = \frac{ \cancel{4}^ { \tiny{2}}}{x} \\ x = 2 \: \: ans[/tex]

In a regression analysis involving 30 observations, the following estimated regressionequation was obtained.y^ =17.6+3.8x 1 −2.3x 2 +7.6x 3 +2.7x 4​For this estimated regression equation SST = 1805 and SSR = 1760. a. At \alpha =α= .05, test the significance of the relationship among the variables.Suppose variables x 1 and x 4 are dropped from the model and the following estimatedregression equation is obtained.y^ =11.1−3.6x 2​ +8.1x 3For this model SST = 1805 and SSR = 1705.b. Compute SSE(x 1 ,x 2 ,x 3 ,x 4 )c. Compute SSE (x2 ,x3 ) d. Use an F test and a .05 level of significance to determine whether x1 and x4 contribute significantly to the model.

Answers

Answer:

(a) There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

(b) [tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

(c) [tex]SSE_{(x_2,x_3)} = 100[/tex]

(d) [tex]x_1[/tex] and [tex]x_4[/tex] are significant

Step-by-step explanation:

Given

[tex]y = 17.6+3.8x_1 - 2.3x_2 +7.6x_3 +2.7x_4[/tex] --- estimated regression equation

[tex]n = 30[/tex]

[tex]p = 4[/tex] --- independent variables i.e. x1 to x4

[tex]SSR = 1760[/tex]

[tex]SST = 1805[/tex]

[tex]\alpha = 0.05[/tex]

Solving (a): Test of significance

We have:

[tex]H_o :[/tex] There is no significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

[tex]H_a :[/tex] There is a significant relationship between y and [tex]x_1, x_2, x_3, x_4[/tex]

First, we calculate the t-score using:

[tex]t = \frac{SSR}{p} \div \frac{SST - SSR}{n - p - 1}[/tex]

[tex]t = \frac{1760}{4} \div \frac{1805- 1760}{30 - 4 - 1}[/tex]

[tex]t = 440 \div \frac{45}{25}[/tex]

[tex]t = 440 \div 1.8[/tex]

[tex]t = 244.44[/tex]

Next, we calculate the p value from the t score

Where:

[tex]df = n - p - 1[/tex]

[tex]df = 30 -4 - 1=25[/tex]

The p value when [tex]t = 244.44[/tex] and [tex]df = 25[/tex] is:

[tex]p =0[/tex]

So:

[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]

Solving (b): [tex]SSE(x_1 ,x_2 ,x_3 ,x_4)[/tex]

To calculate SSE, we use:

[tex]SSE = SST - SSR[/tex]

Given that:

[tex]SSR = 1760[/tex] ----------- [tex](x_1 ,x_2 ,x_3 ,x_4)[/tex]

[tex]SST = 1805[/tex]

So:

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4)} = 1805 - 1760[/tex]

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

Solving (c): [tex]SSE(x_2 ,x_3)[/tex]

To calculate SSE, we use:

[tex]SSE = SST - SSR[/tex]

Given that:

[tex]SSR = 1705[/tex] ----------- [tex](x_2 ,x_3)[/tex]

[tex]SST = 1805[/tex]

So:

[tex]SSE_{(x_2,x_3)} = 1805 - 1705[/tex]

[tex]SSE_{(x_2,x_3)} = 100[/tex]

Solving (d): F test of significance

The null and alternate hypothesis are:

We have:

[tex]H_o :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are not significant

[tex]H_a :[/tex] [tex]x_1[/tex] and [tex]x_4[/tex] are significant

For this model:

[tex]y =11.1 -3.6x_2+8.1x_3[/tex]

[tex]SSE_{(x_2,x_3)} = 100[/tex]

[tex]SST = 1805[/tex]

[tex]SSR_{(x_2 ,x_3)} = 1705[/tex]

[tex]SSE_{(x_1 ,x_2 ,x_3 ,x_4) }= 45[/tex]

[tex]p_{(x_2,x_3)} = 2[/tex]

[tex]\alpha = 0.05[/tex]

Calculate the t-score

[tex]t = \frac{SSE_{(x_2,x_3)}-SSE_{(x_1,x_2,x_3,x_4)}}{p_{(x_2,x_3)}} \div \frac{SSE_{(x_1,x_2,x_3,x_4)}}{n - p - 1}[/tex]

[tex]t = \frac{100-45}{2} \div \frac{45}{30 - 4 - 1}[/tex]

[tex]t = \frac{55}{2} \div \frac{45}{25}[/tex]

[tex]t = 27.5 \div 1.8[/tex]

[tex]t = 15.28[/tex]

Next, we calculate the p value from the t score

Where:

[tex]df = n - p - 1[/tex]

[tex]df = 30 -4 - 1=25[/tex]

The p value when [tex]t = 15.28[/tex] and [tex]df = 25[/tex] is:

[tex]p =0[/tex]

So:

[tex]p < \alpha[/tex] i.e. [tex]0 < 0.05[/tex]

Hence, we reject the null hypothesis

Which expression are greater than 1/2? Choose all the apply

Answers

Answer:

25/30

5/8

Step-by-step explanation:

Which fraction is it out of all of these 6/14,5/8,25/30,or 3/6?

to determine which fractions are greater than 1/2, convert the fractions to decimals

to convert to decimals, divide the numerator by the denominator

1/2 = 0.5  less than half

6/14 = 0.43 less than half

5/8 = 0.625 greater than half

25 / 30 = 0.83 greater than half

3 / 6 = 0.5 equal to half

When comparing two box-plots that show the same type of information, what determines agreement within the data?
A.the range of the quartiles in each data set
B.the median of each data set
C.the mean of each data set
D.the number of values in each data set

Answers

Answer:

c.the mean of each data set

Answer:

A

Step-by-step explanation:

what percentage of the appies are yellow?

Answers

5*6=30
6 yellow apples
6/30=.2
.2*100=20
20%

Answer:

20%

Step-by-step explanation:

6 out of 30. = 1/5 = multiply 5*20= 100 and 1*20= 100 so it is 20% of 100.

You buy a family-size box of laundry detergent that contains 40 cups. , how many loads of wash can you do?

SEE IMAGE BELOW

Answers

Answer:

30 loads

Step-by-step explanation:

You simply divide 40 by 1 1/3 which gives you 30 meaning you can wash 30 loads.

Answer:

30

Step-by-step explanation:

To solve this, you want to do 40÷[tex]1\frac{1}{3}[/tex], or [tex]\frac{40}{1}[/tex]÷[tex]\frac{4}{3}[/tex], which is the same as  [tex]\frac{40}{1}[/tex]×[tex]\frac{3}{4}[/tex].

This can be solved to [tex]\frac{120}{4}[/tex] and simplifies to 30.

**This content involves writing expressions from worded questions and operations with fractions, which you may wish to revise. I'm always happy to help!

How do you solve x[tex]x^{2} +4x+3=0[/tex]?

Answers

Answer:

[tex]{ \tt{ {x}^{2} + 4x + 3 = 0}} \\ { \tt{(x + 1)(x + 3) = 0}} \\ \\ { \tt{x = - 1 \: \: and \: \: - 3}}[/tex]

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.

xy = 2

a. Find dy/dt, when x = 4, given that dx/dt = 13.
b. Find dx/dt, when x = 1, given that dy/dt = -9.

Answers

Answer:

a. [tex]\frac{dy}{dt} = -\frac{13}{8}[/tex]

b. [tex]\frac{dx}{dt} = \frac{9}{2}[/tex]

Step-by-step explanation:

To solve this question, we apply implicit differentiation.

xy = 2

Applying the implicit differentiation:

[tex]y\frac{dx}{dt} + x\frac{dy}{dt} = \frac{d}{dt}(2)[/tex]

[tex]y\frac{dx}{dt} + x\frac{dy}{dt} = 0[/tex]

a. Find dy/dt, when x = 4, given that dx/dt = 13.

x = 4

So

[tex]xy = 2[/tex]

[tex]4y = 2[/tex]

[tex]y = \frac{2}{4} = \frac{1}{2}[/tex]

Then

[tex]y\frac{dx}{dt} + x\frac{dy}{dt} = 0[/tex]

[tex]\frac{1}{2}(13) + 4\frac{dy}{dt} = 0[/tex]

[tex]4\frac{dy}{dt} = -\frac{13}{2}[/tex]

[tex]\frac{dy}{dt} = -\frac{13}{8}[/tex]

b. Find dx/dt, when x = 1, given that dy/dt = -9.

x = 1

So

[tex]xy = 2[/tex]

[tex]y = 2[/tex]

Then

[tex]y\frac{dx}{dt} + x\frac{dy}{dt} = 0[/tex]

[tex]2\frac{dx}{dt} - 9 = 0[/tex]

[tex]2\frac{dx}{dt} = 9[/tex]

[tex]\frac{dx}{dt} = \frac{9}{2}[/tex]

Use the substitution method to solve the system of equations. Choose the correct ordered pair. 4(x + 4) = 8(y + 2); 18y - 22 = 3x + 2​

Answers

x = 30

y = 2

Get the explanation from the image I have shared.

Hope it helps you

Which of the following sets of data does not contain an outlier?
A.16, 17, 20, 19.48
B.59. 60. 61, 67.65
C.95.99.97.94.60
D.-1.2.1.0.5.16

Answers

Answer:

it is a letter b

Step-by-step explanation:

that does not contain an outlet

17.
What is the value of the expression
2a + 5b + 3c for a = 12, b = 6, and
C=3?
A 10
B 21
C49
D 63

Answers

D. 63

2a+5b+3c

2(12)+5(6)+3(3)

24+30+9=63

Hope this helps! :)

ANSWER ASAP IM TIMED 30 POINTS

Which shape below will form a tessellation?

A. regular heptagons

B. regular pentagons

C. regular octagons

D. regular hexagons

Answers

Answer:

Step-by-step explanation:

D ANSWER D

Answer:

d

Step-by-step explanation:

Simplify the expression.

4(11 + 7) ÷ (7 – 5)

Answers

Answer:

36

Step-by-step explanation:

4(11 + 7) ÷ (7 – 5)

4 * 18 ÷ 2

=36

90 dollars ratio in 1:2:3

Answers

Answer:

3:6:9

Step-by-step explanation:

1/1+2+3 × 90 = 15

2/1+2+3 × 90 = 30

3/1+2+3 × 90 = 45

Part A: Factor x2b2 − xb2 − 6b2. Show your work.

Part B: Factor x2 + 4x + 4. Show your work.

Part C: Factor x2 − 4. Show your work

Answers

Answer:

A.2b2(x2-x-3)

B.x2+2x+2x+4

=x(x+2)+2(x+2)

=(x+2)(x+2)

C.x2-2^2

=(x+2)(x+2)

If the first and last terms of a trinomial are perfect squares, the trinomial is a perfect square
trinomial.
True
O False

Answers

Answer:

A trinomial is a perfect trinomial if the first and last terms are positive, perfect squares and the middle term is twice the product of their square roots.

Therefore, if you mean perfect as in positive, the answer is True.

Other Questions
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