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3 years ago

Is 4/6 more than 5/6

Mathematics

2 answers:

Pachacha [2.7K]3 years ago

7 0

4/6 is 4 parts of 6, like 4 pieces of a 6 slice pizza, if you have 5/6 of that pizza, then 5/6 is more than 4/6

inysia [295]3 years ago

5 0

Im pretty sure it is
Good Luck
hope i helped!!!!

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F(x)=x^2+10x-12

weqwewe [10]

Answer:

Therefore the two values required are 2.08 and -11.08.

Step-by-step explanation:

i) f(x) = $x^{2}$ + 10x - 12 is a quadratic equation and a quadratic equation will have two roots whose values when substituted in to the given quadratic equation will give the value.

ii) The question is therefore essentially asking us to find the roots of the given quadratic equation.

This can be done by equating the given quadratic equation to zero and then solving the equation for its roots.

∴ $x^{2}$ + 10x -12 = 0 ⇒ x = (-10 ± $\sqrt{10^{2} +(48)}$ ) ÷ 2

= (-10 ± 12.166)÷2 = 2.08, -11.08

Therefore the two values required are 2.08 and -11.08

3 0

3 years ago

All of the following expressions are equivalent except _____.

defon

Answer:

3. x-2

Step-by-step explanation:

Let's start with 1. we know that 2-x is the same as -2+x and since addition can be written as -2+x or x+(-2) and equal the same thing, we know that x-2 is the one that is not equivalent.

4 0

3 years ago

A typical basketball player scores on 70% of free throw attempts. During a basketball practice, Lebron james attempted 40 free t

avanturin [10]

The answer is 30
40 times 3/4 is 30
40 times .75 is also 30

5 0

2 years ago

Evaluate the expression for the given value<br><br> 11 + r r = 7

Cerrena [4.2K]

$11+r =7$

r=-4

6 0

2 years ago

Read 2 more answers

Find the integral <br> ∫√(9+x)/(9-x)

densk [106]

I suppose you mean

$\displaystyle \int \frac{\sqrt{9+x}}{9-x} \, dx$

Substitute y = √(9 + x). Solving for x gives x = y² - 9, so that 9 - x = 18 - y², and we have differential dx = 2y dy. Replacing everything in the integral gives

$\displaystyle \int \frac{2y^2}{18 - y^2} \, dy$

Simplify the integrand by dividing:

$\dfrac{2y^2}{18 - y^2} = -2 + \dfrac{36}{18 - y^2}$

$\implies \displaystyle \int \left(\frac{36}{18-y^2} - 2\right) \, dy$

For the first term of this new integral, we have the partial fraction expansion

$\dfrac1{18 - y^2} = \dfrac1{\sqrt{72}} \left(\dfrac1{\sqrt{18}-y} + \dfrac1{\sqrt{18}+y}\right)$

$\implies \displaystyle \frac{36}{\sqrt{72}} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

The rest is trivial:

$\displaystyle \sqrt{18} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

$= \displaystyle \sqrt{18} \left(\ln\left|\sqrt{18}+y\right| - \ln\left|\sqrt{18}-y\right|\right) - 2y + C$

$= \displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+y}{\sqrt{18}-y}\right| - 2y + C$

$= \boxed{\displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+\sqrt{9+x}}{\sqrt{18}-\sqrt{9+x}}\right| - 2\sqrt{9+x} + C}$

6 0

2 years ago