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3 years ago

I need help with this

Mathematics

2 answers:

Verizon [17]3 years ago

8 0

b. 8 x 10³

hope this helps

DochEvi [55]3 years ago

4 0

Answer: B. 8 x 10^3

Step-by-step explanation:

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Allison has three containers with 25 crayons in each. She also has four boxes of markers with 12 markers in each box. She gets 1

Advocard [28]

25x3=75
4x12=48
48-10=38

75 crayons and 38 markers

3 0

3 years ago

Which number lint represents the solution set for the inequity -1/2x>4

g100num [7]

Answer:

B. x < -8

Step-by-step explanation:

Well first we need to get x by itself.

To do that we do 4 / -1/2 = -8

And since a negative was divided the > changes to a <.

So x<-8.

If x is less than -8 the line starts at -8 and goes to the left.

Thus,

the answer is choice b.

Hope this helps :)

5 0

2 years ago

Please write blue splat

ludmilkaskok [199]

Answer:

Blue splat

Step-by-step explanation:

Blue splat

7 0

3 years ago

WILL MARK BRAINLIEST!!!!!!MUST BE CORRECT!

ExtremeBDS [4]

Hi there! The answer is 9 square units

To find the area of a triangle we use the following formula:
Area = 0.5 * base * height.

The base of this triangle:
AC = 6

The height of this triangle:
BD = 3

Filling in this formula gives us:
Area = 0.5 * 3 * 6 = 9

Therefore, the answer is 9 square units

7 0

2 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