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3 years ago

When checking you answer to an addition or subraction problem, use the _______ operation? OPPOSITE

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

It is the opposite. There is nothing called same operation...

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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

Estimate 59 times 8 A. 60 B. 480 C. 700 D. 240

Lana71 [14]

Answer:

480 is the answer

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Jan conducts an experiment and tosses a coin 40 times. She gets heads 30 times instead of the expected 20. What could be a reaso

Nady [450]

the answer is B. hope i helped

8 0

3 years ago

Which of these cannot be an 2-value? A. -0.88 B. O C. 0.88 D. 1

defon

Answer:

Step-by-step explanation:

6 0

3 years ago

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Oh no! Thomas is lost in the GGBL building after an exam, and trying to get out. Thereare four identical hallways leading away f

Brums [2.3K]

Answer:

Check the explanation

Step-by-step explanation:

Since each and every one of the 4 hallways is equally likely then

$P(H1)=1/4P(H2)=1/2P(H3)=1/4$

Now, if he chooses H1 he escapes after 12 minutes then

$E(T|H1)=12$

If he chooses H2 then he wastes 10 minutes and then he is again in the same starting

position so he expects to escape in E(T) minutes, then

$E(T|H2)=10+E(T)$

Analogously, if he chooses H3 then he wastes 40 minutes and then he is again in the same starting

position so he expects to escape in E(T) minutes, then

$E(T|H3)=40+E(T)$

Therefore

$E(T) = E(T|H1)P(H1) + E(T|H2)P(H2) + E(T H3)P(H3) = (12) + (E(T) +10) + (E(T) + 40) = 3+ *E(T)+5+ 10 = E(T) +18 CON A'$

So we have

$E(T) = E(T) + 18 E(T) = 18 E(T) = 4(18) = 72$

Therefore he is expected to escape in 72 minutes

7 0

3 years ago