What percent of 1/2 is 2/5

An igneous rock may change into which of the following?

grin007 [14]

By going through the rock cycle igneous rock can change into any other type of rock, so A. any of the above

3 0

3 years ago

Read 2 more answers

Solve.<br> Subtract 6x2 – X – 7 from the product of x2 + x + 5 and x - 6.

tresset_1 [31]

Answer:

x ^ 3 − 11 x ^ 2 + X - x − 23

Step-by-step explanation:

you have to Simplify the expression. Hope that helped!! :)

PLEASE MARK BRAINLIEST!!

8 0

3 years ago

In order for RECT to be a rectangle, what must the value of x be?

Readme [11.4K]

Answer:

x = 21

Step-by-step explanation:

If RECT is a rectangle, each of its interior angles are right angles, 90°. Since ∠RTC is split into two angles, the sum of the two angles must be 90°. (Angles that add to 90° are called complementary angles).

∠RTC = ∠RTE + ∠ETC Formula for complementary angles

90 = 2x + 6 + 42 Substitute the angles

90 = 2x + 48 Combined like terms to simplify

Now isolate "x" to solve

90 - 48 = 2x + 48 - 48 Subtract 48 from both sides

42 = 2x

42/2 = 2x/2 Divide both sides by 2

x = 21

Therefore x is 21.

6 0

3 years ago

Someone please help out

exis [7]

Answer:

$39+0.15m \leqslant 50$

Step-by-step explanation:

From the question, Diana wants her total bill to be less than $50.

Diana pays a fixed monthly bill of

$39 for her mobile phone and an additional

$0.15 per minute for phone internet service.

The bill Diana pays for m minutes of using the internet is $0.15m in a month.

The total monthly bill becomes,

$39+0.15m$

The maximum bill she wants to pay is $50.

$\implies39+0.15m \leqslant 50$

This inequality can be solved to determine the number of available minutes.

4 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