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2 years ago

Please help with maths!!!

Mathematics

1 answer:

zimovet [89]2 years ago

8 0

It is the angle where it says "A".

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What is the slope of the line that passes through the points (10,3 and (7,0)

user100 [1]

Answer:

-3/-3=1

Step-by-step explanation:

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1 year ago

Please help me out!

babunello [35]

Answer:

9.8

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Rewrite using only positive exponents. <br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3a%20%5E%7B%20-%201%7D%20b%

Komok [63]

That would be a / 3b^2

(re call that 1 /a^-1 = a)

6 0

2 years ago

1.What is the area of this parallelogram?

dexar [7]

Question 1: Option D

Area of the parallelogram = 132 cm²

Question 2: Option B

Area of the parallelogram = 3.60 ft²

Solution:

Question 1:

Base of the parallelogram = 6 + 5 = 11 cm

Height of the parallelogram = 12 cm

Area of the parallelogram = Base × Height

= 11 × 12

Area of the parallelogram = 132 cm²

Option D is the correct answer.

Question 2:

Base of the parallelogram = 1.9 + 0.5 = 2.4 ft

Height of the parallelogram = 1.5 ft

Area of the parallelogram = Base × Height

= 2.4 × 1.5

Area of the parallelogram = 3.60 ft²

Option B is the correct answer.

7 0

3 years ago

In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin

Mademuasel [1]

Answer: $7\frac{5}{7}$ unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

And, ∠ PAB ≅ ∠ PBA

∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

$\triangle PDC \sim \triangle PAB$

Thus, By the definition of similarity,

$\frac{PD}{PA} = \frac{DC}{AB}$

$\frac{PD}{PD+3} = \frac{11}{18}$ ( because PA = PD+DA)

⇒ 18 PD = 11 PD +33

⇒7PD = 33

⇒ PD = 33/7

Again by the definition of similarity,

$\frac{PC}{PB} = \frac{DC}{AB}$

$\frac{PC}{PC+7} = \frac{11}{18}$ ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA = PD+DA = 33/7 + 3 = $7\frac{5}{7}$

And, PB = PC + CB = 11 + 7 = 18

8 0

2 years ago