All categories

2 years ago

HELP PLEASE!!!! MULTIPLE CHOICE!!!

Mathematics

1 answer:

kirill115 [55]2 years ago

5 0

Answer:

1.) a standard form and general equation

2.) 8

Step-by-step explanation:

The conic forms have a standard form and general equation. Conic section is the intersection in the plane. The slope of intersecting plane is greater than the slope of the cone.

Latus Ractum of a parabola is the curve through the focus which is parallel to the directrix. Half latus rectum is called semilatus rectum. It is the chord which passes perpendicular to the principal axis of ellipse, parabola and hyperbola.

You might be interested in

Lacy has a $1000 bond with a 6.2% coupon that she purchased for $1025. What is the yield of this bond? Round your answer to 2 de

Ede4ka [16]

So the question is asking what is 6.2% of 1000? Then rounding our decimals?

4 0

2 years ago

Read 2 more answers

One-fourth of a

Whitepunk [10]

Answer:

24 bowls

Step-by-step explanation:

Its 24 bowls because 1/4 fits into one bowl. 4 bags of popcorn with fit in 4 bowls. So then 6 times 4 equals 24.

3 0

2 years ago

Write the following expression in simplest form: 2y+X+4+7x+3y+4*

tiny-mole [99]

2y+X+4+7x+3y+4* in simplest form is:
8x+5y+8

7 0

3 years ago

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

Help quiz question!!!

SVEN [57.7K]

Answer:301.59389

Step-by-step explanation:

254.47+47.12389=301.59389

4 0

2 years ago