All categories

3 years ago

Will some one help me please will friend

Mathematics

1 answer:

aliina [53]3 years ago

7 0

Add the equations,just the way they appear there.

-- Add the top 'x' to the bottom 'x'. Write the sum under the 'x's.
-- Add the '+y' and the '-y'. Write the sum under the 'y's.
-- Add the '2' and the '4'. Write the sum under them, with n " = " sign
before it.

You should now have an equation with only 'x' in it and no 'y'.
You can easily solve that one and find out the value of 'x'.

Once you know the value of 'x', go back to either one of the original
equations, and plug the number-value of 'x' in place of 'x'.

You'll then have an equation with only 'y' in it and no 'x'.
You can easily solve that one and find out the value of 'y'.

You might be interested in

Last week 24,000 fans attended a football match. this week three times as many bought tickets, but one-sixth of them cancelled t

lukranit [14]

Answer:

C. 60,000

Step-by-step explanation:

Given

24,000 fans attended a football match

three times as many bought tickets

one-sixth of them cancelled their tickets

Three times as many bought tickets

24,000 * 3 = 72,000

one-sixth of them cancelled their tickets

72,000 * 1/6 = 12,000

Subtract

72,000 - 12,000 = 60,000

Answer

60,000 people are attending this week

6 0

4 years ago

Please help mee pleaseee

nalin [4]

Answer:

The exact answer in terms of radicals is $x = 5*\sqrt[3]{25}$

The approximate answer is $x \approx 14.62009$ (accurate to 5 decimal places)

===============================================

Work Shown:

Let $y = \sqrt[5]{x^3}$

So the equation reduces to -7 = 8-3y

Let's solve for y

-7 = 8-3y

8-3y = -7

-3y = -7-8 ... subtract 8 from both sides

-3y = -15

y = -15/(-3) ... divide both sides by -3

y = 5

-----------

Since $y = \sqrt[5]{x^3}$ and y = 5, this means we can equate the two expressions and solve for x

$y = 5$

$\sqrt[5]{x^3} = 5$

$x^3 = 5^5$ Raise both sides to the 5th power

$x^3 = 3125$

$x = \sqrt[3]{3125}$ Apply cube root to both sides

$x = \sqrt[3]{125*25}$

$x = \sqrt[3]{125}*\sqrt[3]{25}$

$x = \sqrt[3]{5^3}*\sqrt[3]{25}$

$x = 5*\sqrt[3]{25}$

$x \approx 14.62009$

4 0

3 years ago

Read 2 more answers

Using n for the variable, write an algebraic expression that represents a number that is divisible by the given number. (Hint: A

iris [78.8K]

Answer:

'5n' is the correct answer.

Step-by-step explanation:

Let 'n' be any integer i.e. a number from the set {....., -3,-2,-1,0,1,2,3, ..... }

so 'n' can be termed as the variable here because its value can change and can be any value from the above set.

A number 'q' that can be divided by a a given number 'p' can be written as:

$n \times p$

When divided by 'p' :

$\dfrac{q}p = \dfrac{n \times p}{p}\\\dfrac{q}p = n$

So, The number 'q' is completely divisible by 'p' leaving 'n' as the quotient.

Using this concept, let us solve the questions:

a) Using 'n' as the variable, a number that is divisible by 5 can be written as:

$5n$

Read more on Brainly.com - brainly.com/question/16919676#readmore

7 0

3 years ago

Ví dụ 1.5. Một người đi mua hàng 3 lần. Xác suất lần đầu mua được hàng tốt là 0,7.

vodka [1.7K]

Answer:

a) 0.50575,

b) 0.042

Step-by-step explanation:

Example 1.5. A person goes shopping 3 times. The probability of buying a good product for the first time is 0.7.

If the first time you can buy good products, the next time you can buy good products is 0.85; (I interpret this as, if you buy a good product, then the next time you buy a good product is 0.85).

And if the last time I bought a bad product, the next time I bought a good one is 0.6. Calculate the probability that:

a) All three times the person bought good goods.

P(Good on 1st shopping event AND Good on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Good on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st and 2nd shopping events yield Good) =

(0.7)(0.85)(0.85) =

0.50575

b) Only the second time that person buys a bad product.

P(Good on 1st shopping event AND Bad on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Bad on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st is Good and 2nd is Bad shopping events) =

(0.7)(1-0.85)(1-0.6) =

(0.7)(0.15)(0.4) =

0.042

5 0

3 years ago

If a plate is moving 3.3 cm/yr, what distance [in kilometers (km)] will the plate move in 1 million years?

sasho [114]

$\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}$

3.3cm = 0.000033km (divide by 100,000.

So, 0.000033km/yr.

0.000033 × 1,000,000

=33km

Thanks
Hope it helps

3 0

3 years ago