All categories

3 years ago

R over 10 + 7 = 8 please help? two step equations

2 answers:

7 0

My guess

$\frac{r}{10 + 7 = 8}$
1. Add the numbers
10+7=17

2. Check the equality
17=8

The equality is false because both sides are different

lisabon 2012 [21]3 years ago

3 0

Take away the zero. Add 7 to 1 and u get 8. The reason u would take away 0 because u don’t count it in this problem.

You might be interested in

You are putting together a single elimination tournament. You have 4368 people sign up to compete. How many game will have to be

frosja888 [35]

Answer:

you will have to have 2,184 games to have a winner

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Expand and simplify (n-2)(3n+7) + 2n(2n+3)

KonstantinChe [14]

3n^2+7n-6n-14+4n^2+6n= 7n^2+7n-14

Hope that helps and best of wishes!

6 0

3 years ago

Convert 0.34285714285 into an improper fraction

Stolb23 [73]

Answer:

34285714285/100000000000

Step-by-step explanation:

To write 0.34285714285 as a fraction you have to write 0.34285714285 as the numerator and put 1 as the denominator. Now you multiply the numerator and denominator by a number that makes the numerator to a whole number.

And finally, we have:

0.34285714285 as a fraction equals 34285714285/100000000000

5 0

3 years ago

On a backpacking trip through Europe, Troy spent 20 Swiss francs to stay one night in a youth hostel. If the foreign exchange ra

g100num [7]

Troy spent $9.09 on his backpacking trip to Europe

3 0

3 years ago

Read 2 more answers

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

2 years ago