All categories

3 years ago

Anyone can help me solve this question3a^2b^2-17abxy+10x^y^2 using the Cross multiplication method!?

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:\frac{5}{32a^{13}b}

Step-by-step explanation:

You might be interested in

Factor the expression over the complex numbers. x3+20x−4x2−80

sveta [45]

Answer: (x^2+20)(x-4)

Step-by-step explanation:

See attached picture

3 0

3 years ago

In the 1970s, due to world events, there was a gasoline shortage in the United States. There were often long lines of cars wai

barxatty [35]

Answer:

Part A: The average car length is 9.1 feet to the nearest tenth foot

Part B: The line would be 9100 feet to contained 1000 cars

Step-by-step explanation:

* Lets explain how to solve the problem

# Part A:

- There were 62 cars in a line that stretched 567 feet

- The cars are lined up bumper-to-bumper

- That means there is no empty spaces between the cars

* To find the average length of the car we will divide the length of

the line by the numbers of the cars

∵ The average car length = length of the line/number of cars

∵ The length of the line is 567 feet

∵ The numbers of the cars is 62 cars

∴ The average car length = 567/62 = 9.1 feet

* The average car length is 9.1 feet to the nearest tenth foot

# Part B:

- There are 1000 cars

- We need to find the length of line which contained the cars

∵ The average car length = length of the line/number of cars

∵ The average of car length is 9.1 feet

∵ The number of the cars is 1000 cars

∴ 9.1 = length of the line/1000

- Multiply both sides by 1000

∴ The length of the line = 9.1 × 1000 = 9100 feet

∴ The line would be 9100 feet to contained 1000 cars

7 0

3 years ago

What does BC equal ?

Mariulka [41]

Answer:

Step-by-step explanation:

bc = df x 2

14 x 2 = 28

I hope this helps

4 0

3 years ago

In triangle ABC, AD is a median. FE is a straight line parallel to BC cutting the remaining sides AB at F and AC at E and cuttin

daser333 [38]

Answer:

Step-by-step explanation:

since AD is a median it implies that triangle ABC is bisected to two equal right angled triangle which are ADB and ADC.

FE is parrallel to BC and cuts AB at F and AC at E shows that there are two similar triangles formed which are AFE and ABC.

Recall that ADC is a right angled triangle, ED bisects a right angled triangle the the ADE = $45^{o}$ .

Now, Let FD bisect angle ADB,

then ADF = $45^{o}$ too.

Since AFX is similar to Triangle ABD and that Triangle AEX is similar to Triangle ACD, then EDX is similar to FDX

FDE = ADF + ADE = $90^{o}$

4 0

3 years ago

Will give brainliest!

miv72 [106K]

Answer:

it means that the roots of the quadratic equation is real and distinct and that means that 40 is a positive value

6 0

2 years ago