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4 years ago

Need help with this work sheet

Mathematics

1 answer:

Lesechka [4]4 years ago

5 0

Answer:

its alot to explain but i will try to make it as simple as possible

Step-by-step explanation:

your first goal is to make each problem into the form ax^2+bx+c=0

number 1, 2, 7 and 8 is already done for you

now all you have to do is plug in each number in the standard form into the quadtratic formula.

basically at this point you can just use your calculator to do the rest of the work. dont forget parentheses so it doesnt get confused...

or you can perform the algebraic work.. its all just a matter of plugging in the right numbers into the quadratic formula...

cant really do the work for you since im on my phone. but yeah all you need to do step one is transform each problem into ax^2+bx+c=0 form

then step 2, plug in each number in to the quadtratic formula. from there calculate using basic algebraic rules

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vesna_86 [32]

Answer:

Step-by-step explanation:

x^2/y

Let x=10 and y=5

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3 years ago

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KengaRu [80]

Answer:

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4 years ago

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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

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