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spin [16.1K]
3 years ago
6

Helppppppppppppppppp

Mathematics
2 answers:
lara31 [8.8K]3 years ago
6 0
Angle 1 and angle 8 are congruent
sashaice [31]3 years ago
6 0

Answer:

Angle 1 and angle 8 are congruent

Hope this helped have an amazing day!

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37 Ifx - 7 = 1, then x + 19​
Virty [35]

Answer:

152

Step-by-step explanation:

x =7+1 = 8

8 x 19 =152

3 0
3 years ago
Read 2 more answers
Cleo added 3a+4b and got 7ab. Three of these statements explain why her answer is wrong. Which one does not
Triss [41]

Answer:

3 and 4 are factors of 12 so they should be multiplied to get a product of 12 ab. (Answer D)

Step-by-step explanation:

You do not multiply in addition unless an exponent is present. You should not multiply or add considering these are not like terms. Therefore, D is wrong.

5 0
3 years ago
Angles H and F are corresponding angles<br> True or False?
lutik1710 [3]
True! they are corresponding angles.
5 0
2 years ago
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NO LINKS. Find the measure of the angle or arc indicated. Part 2<br>​
sergeinik [125]

9514 1404 393

Answer:

  14.  C) 136°

  15.  C) 40°

Step-by-step explanation:

Inscribed angles are half the measure of the arc they intercept. For an inscribed quadrilateral, this means opposite angles are supplementary.

__

14) ∠H +∠W = 180°

  34x +55x +2 = 180

  89x = 178 . . . . . . . . . subtract 2

  x = 2 . . . . . . . . . . . . . divide by 89

  arc VX = 2(34x) = 68(2) = 136 . . . degrees

__

15) The sum of angles in the triangle is 180°.

  ? + 80° + (120°/2) = 180°

  ? = 40° . . . . . . . . . . subtract 140°

6 0
3 years ago
Read 2 more answers
An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

7 0
3 years ago
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