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2 years ago

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jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer: jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj its 8

Step-by-step explanation:

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Emma needs to divide 7 by 1/2. Which strategy can Emma use to get the answer?

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14

Step-by-step explanation:

Convert the fraction into a decimal

1/2 = 0.5

Divide

7/0.5 = 14

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On a coordinate plane, 2 trapezoids are shown. Trapezoid A B C D has points (negative 2, 4), (2, 4), (negative 3, 0) and (3, 0).

devlian [24]

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2 years ago

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gavmur [86]

Answer:

m∠QPM=43°

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠NPQ=m∠MPN+m∠MPQ

we have

m∠NPQ=(9x-25)°

m∠MPN=(4x+12)°

m∠MPQ=(3x-5)°

substitute the given values and solve for x

(9x-25)°=(4x+12)°+(3x-5)°

(9x-25)°=(7x+7)°

9x-7x=25+7

2x=32

x=16

Find the measure of angle QPM

Remember that

m∠QPM=m∠MPQ

m∠MPQ=(3x-5)°

substitute the value of x

m∠MPQ=(3(16)-5)=43°

therefore

m∠QPM=43°

7 0

2 years ago

In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t

Sergeeva-Olga [200]

Answer:

Perimeter = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

$MB = \frac{a}{2}$

Now in ΔAMB , By using Pythagoras theorem

$AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}$

Now, in ΔBMP :

$sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}$

Perimeter of ABM = AB + PB + PM + AM

$\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a$

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0

2 years ago