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mash [69]
3 years ago
6

HELP I NEED HELP ASAP

Mathematics
1 answer:
Allisa [31]3 years ago
5 0

Answer:

3x^2-6x+3=0

x^2-2x+1=0

x^2-x-x+1=0

x(x-1)-(x-1)=0

(x-1)(x-1)=0

x=1,1

so x has only one answer

so it intersects only one time

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A regular pentagon is shown. A regular pentagon has side lengths of 9.4 centimeters and a radius of 8 centimeters. What is the l
Angelina_Jolie [31]

Answer:

6.5 cm

Step-by-step explanation:

The computation of the length of the apothem is shown below:

Given that

The side length is 9.4 centimeters

And, the radius is 8 centimeters

Now based on the above information

As per the attached figure  

AB = 8 cm

BC = 9.4 ÷ 2 = 4.7 cm

Now apply the pythagoras theorem

AB^2 = BC^2 + AC^2

8^2 = 4.7^2 + AC^2

AC^2 = 41.91

AC = 6.47 cm

= 6.5 cm

7 0
3 years ago
What is the area of the parallelogram?
Genrish500 [490]
Equals the product of one of its sides
3 0
2 years ago
Read 2 more answers
Candice is preparing for her final exam in Statistics. She knows she needs an 74 out of 100 to earn an A overall in the course.
Salsk061 [2.6K]

Answer:

Hence the correct option is option b) Yes, the upper level of one standard deviation is 72.

A score of 74 is not within one standard deviation of the mean.

Step-by-step explanation:

Here the given details are,

Mean = 68  

SD = 4  

Distribution is normal.  

Z-score for x = 74 is given as below:  

Z = (X - mean)/SD\\Z = (74 - 68)/4\\Z = 1.5  

So, the score of 74 is 1.5 standard deviations from the mean.  

Mean + 1\timesSD = 68 + 1\times4 = 72Mean - 1\timesSD = 68 - 1\times4 = 64  

Therefore the score is not lies between 64 and 72.  

Yes, the upper level of one standard deviation is 72.  

3 0
3 years ago
Unit 8 right triangles and trigonometry homework 3 similar right triangles and geometric mean
Ugo [173]

The right triangles that have an altitude which forms two right triangles

are similar to the two right triangles formed.

Responses:

1. ΔLJK ~ ΔKJM

ΔLJK ~ ΔLKM

ΔKJM ~ ΔLKM

2. ΔYWZ ~ ΔZWX

ΔYWZ ~ ΔYZW

ΔZWX ~ ΔYZW

3. x = <u>4.8</u>

4. x ≈ <u>14.48</u>

5. x ≈ <u>11.37</u>

6. G.M. = <u>12·√3</u>

7. G.M. = <u>6·√5</u>

<u />

<h3>What condition guarantees the similarity of the right triangles?</h3>

1. ∠LMK = 90° given

∠JMK + ∠LMK  = 180° linear pair angles

∠JMK = 180° - 90° = 90°

∠JKL ≅ ∠JMK All 90° angles are congruent

∠LJK ≅ ∠LJK reflexive property

  • <u>ΔLJK is similar to ΔKJM</u> by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

  • <u>ΔLJK is similar to ΔLKM</u> by AA similarity

By the property of equality for triangles that have equal interior angles, we have;

  • <u>ΔKJM ~ ΔLKM</u>

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

  • <u>ΔYWZ is similar to ΔZWX</u>, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

  • <u>ΔYWZ is similar to ΔYZW</u> by AA similarity postulate

Therefore;

  • <u>ΔZWX ~ ΔYZW</u>

3. The ratio of corresponding sides in similar triangles are equal

From the similar triangles, we have;

\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}

8 × 6 = 10 × x

48 = 10·x

  • x = \dfrac{48}{10} = \underline{4.8}

3. From the similar triangles, we have;

\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}

20 × 21 = x × 29

420 = 29·x

  • x = \dfrac{420}{29 } \approx \underline{14.48}

4. From the similar triangles, we have;

\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}

20 × 48 = 52 × x

  • x = \dfrac{20 \times 48}{52}  = \dfrac{240}{13}  \approx \underline{18.46}

5. From the similar triangles, we have;

\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}

13.2 × 22.4 = 26 × x

  • x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}

6. The geometric mean, G.M. is given by the formula;

G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3  ... x_n}}

The geometric mean of 16 and 27 is therefore;

  • G.M. = \sqrt[2]{16 \times 27}  = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}

  • The geometric mean of 16 and 27 is <u>12·√3</u>

<u />

7. The geometric mean of 5 and 36 is found as follows;

G.M. = \sqrt[2]{5 \times 36}  = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}

  • The geometric mean of 5 and 36 is <u>6·√5</u>

Learn more about the AA similarity postulate and geometric mean here:

brainly.com/question/12002948

brainly.com/question/12457640

7 0
2 years ago
PLEASE HELP!! how do I prove that vertical angles are congruent???
grin007 [14]

Answer:

a b

Step-by-step explanation:

5 0
3 years ago
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