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3 years ago

The length of a rectangle is 4m more than the width.The area of the rectangle is 45 m^2.find the length

Mathematics

1 answer:

Nataly_w [17]3 years ago

4 0

The answer is 9m.

Because 9 - 5 = 4 (difference) and (9)(5) = 45.

Done by trial and error.

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Can i get some help please

Brilliant_brown [7]

Answer:

Step-by-step explanation:

We can use the pythagorean theorem to find the leg.

10^2+x=13^2

x=12

We now find the area given the height of the triangle.

10 X 12=120

120/2=60

3 0

3 years ago

Read 2 more answers

Need help 25 points

Zina [86]

Answer:

The answer is 50.27 yards.

Step-by-step explanation:

You divide the diameter in 2, which gives you the radius of 4. Then you must square the radius, or multiply it twice, 4 times 4, to get 16. Then multiply that by pi or 3.14, which is 50.27. Hope this helped and please give brainliest if possible! :)

3 0

3 years ago

The formula for electrical power (P, in watts) is where V is the voltage in volts and R is the resistance in ohms. For a particu

Firlakuza [10]

Answer:v=√(pr) 4volts

Step-by-step explanation:

v=√(pr)

v=√(32×0.5)

v=√(16)

v=4volts

4 0

4 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

3 years ago

The club has only $475 in its bank account. Write

larisa86 [58]

An inequality that represents this equation would be:

x ≤ 475

They cant spend more than $475, so they can spend less than or $475 as represented in the question.

8 0

2 years ago