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3 years ago

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Mathematics

2 answers:

Mila [183]3 years ago

5 0

Answer:

the answer is D.

!!!!!!!!!!!!!!!!!!!!!!!!

nikitadnepr [17]3 years ago

3 0

Answer:

D, $10i\sqrt{2}$

Step-by-step explanation:

Separate -1, 2, and 100 from each other inside the radical. You can take out -1 as <em>i</em>, and 100 as 10. You now have 10<em>i</em> on the outside of the radical and 2 on the inside. Hope this helps.

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ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago

What is the lateral area of a regular square pyramid if the base edges are of length 24 and perpendicular height is 5?

iren2701 [21]

Answer:

The lateral area is 624 unit²

Step-by-step explanation:

* Lets explain how to solve the problem

- The regular square pyramid has a square base and four congruent

triangles

- The slant height of it = $\sqrt{(\frac{1}{2}b)^{2}+h^{2}}$ , where

b is the length of its base and h is the perpendicular height

- Its lateral area = $\frac{1}{2}.p.l$ , p is the perimeter of the base

and l is the slant height

* Lets solve the problem

∵ The base of the pyramid is a square with side length 24 units

∵ Its perpendicular height is 5 units

∵ The slant height (l) = $\sqrt{(\frac{1}{2}b)^{2}+h^{2}}$

∴ l = The slant height of it = $\sqrt{(\frac{1}{2}.24)^{2}+5^{2}}$

∴ l = $\sqrt{(12)^{2}+25}=\sqrt{144+25}=\sqrt{169}=13$

∴ l = 13 units

∵ Perimeter of the square = b × 4

∴ The perimeter of the base (p) = 24 × 4 = 96 units

∵ The lateral area = $\frac{1}{2}.p.l$

∴ The lateral area = $\frac{1}{2}.(96).(13)$

∴ The lateral area = 624 unit²

* The lateral area is 624 unit²

4 0

3 years ago

Answer this please tell me if I’m correct

jek_recluse [69]

Yes you are correct, :)

6 0

3 years ago

An equation has solutions of m= -5 and m=9. Which would be the equation?

Aliun [14]

Answer:

There could be many but one of them is

(m+5)(m-9)=0....

7 0

3 years ago

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HELP ASAP question 5!!!

AnnyKZ [126]

Step-by-step explanation:

Total area of merged figure = area of vertical rectangle + area of horizontal rectangle.

$\therefore \: 3x(2x - 7) + (3x + 4)x = 85 \\ \\ \therefore \: 6 {x}^{2} - 21x+ 3 {x}^{2} + 4x = 85 \\ \\ \therefore \: 6 {x}^{2} + 3 {x}^{2} - 21x + 4x - 85 = 0 \\ \\ \huge \red {\boxed {\therefore \: 9 {x}^{2} - 17x - 85 = 0}} \\ hence \: proved.$

8 0

3 years ago

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