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3 years ago

Write and evaluate the expression. Then, select the correct answer.

Mathematics

1 answer:

Law Incorporation [45]3 years ago

4 0

Answer:

Step-by-step explanation: So we need to write and solve an equation to solve for m, which is 12. Simply plug in 12 for m in the equation.

Step 1: 2-x/4+10

Step 2: 2-12/4+10

Step 3: 2-3+10

Step 4: -1+10=9

Plug 12 in for x

12 divided by 4 is 3 and 2-3 is -1

-1 + 10 is 9 -1 + 10 is 9

i hop it help u

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Need helpp fastt plss will give brainliest to who is correct plss helpp

Juliette [100K]

Answer:

12(3+1)

Step-by-step explanation:

use distributive property to solve

13 times 3 is 36

12 times 1 is 12

so 12(3+1) is equal to 36+12

have a good day :)

8 0

2 years ago

Read 2 more answers

a fruit market sold 3,849 apples 3,498 oranges,and 3894 pears in one day.round each number to the nearest hundred.then order the

Ainat [17]

First is pears ,next is apple, and the last is oranges. 3,900, 3800, 3500. I hope that this helps.

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3 years ago

A man who is 6 feet tall casts a shadow that is 11 feet long. At the same time, a tree casts a shadow that is 33 feet long. What

Rina8888 [55]

Set up a proportion:

6 feet over 11 feet long equals x feet over 33 feet long

6/11 = x/33

198 = 11x

x = 18

The height of the tree is 18 feet tall

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2 years ago

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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

Use a graphing calculator or online application to find the solutions to x^2-9 =6sqrt3/x to the nearest tenth.

Olegator [25]

Answer:

{-1.7, 3.5}

Step-by-step explanation:

I find it convenient to graph the difference of the two sides of the equation. That difference is zero when the equation is satisfied. This equation has two solutions, near x = -1.7 and x = 3.5.

4 0

2 years ago