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4 years ago

What is expanded form for 4,408,730?

Mathematics

1 answer:

olganol [36]4 years ago

7 0

4,000,000
400,000
00,000
8,000
700
30
0

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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5

Gnesinka [82]

$\large\underline{\sf{Solution-}}$

Given that,

In triangle TPQ, 

RS || PQ,

RS = 3 cm,

PQ = 6 cm,

ar(∆ TRS) = 15 sq. cm

As it is given that, RS || PQ

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

$\rm\implies \: \triangle TPQ \: \sim \: \triangle TRS \: \: \: \: \: \: \{AA \}$

Now, We know 

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)} = \dfrac{ {PQ}^{2} }{ {RS}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{ {6}^{2} }{ {3}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{36 }{9}$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = 4$

$\rm\implies \:ar( \triangle \: TPQ) = 60 \: {cm}^{2}$

3 0

2 years ago

15 points with easy explanation please

GaryK [48]

Answer:

that would be 38 degrees

Step-by-step explanation:

7 0

3 years ago

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15% of the toddlers in a preschool class drink water with their lunch. How many toddlers are in the class if 3 drink water with

Naddika [18.5K]

15%=3
5%=1
5(20)=100
1(20)=20

So, there are 20 students in the class.

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3 years ago

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The ratio of the angles in a triangle is 3:10:7. what is the measure of the smallest angle?

STALIN [3.7K]

So add all of them and then see smallest one sorry if I'm wrong

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4 years ago

What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27

malfutka [58]

Answer with explanation:

$\rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}$

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

=(y+2)(y+3)

→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)

6 0

3 years ago

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