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3 years ago

The quadratic x? - x = 2 has two solutions. What is the larger of the two solutions?

1 answer:

5 0

I believe the 2 is the largest

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ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

The slope of a line given two points $(x_0,y_0)$ and $(x_1,y_1)$ is

$m=\frac{y_1-y_0}{x_1-x_0}=\frac{15-(-12)}{18-2}=\frac{27}{16}$

7 0

3 years ago

nevsk [136]

Answer:

1,500,000,000

Step-by-step explanation:

6 0

2 years ago

DochEvi [55]

So the rule with multiplying exponents of the same base is $x^m*x^n=x^{m+n}$ . Apply this rule here:

$7^{-\frac{5}{6}}*7^{-\frac{7}{6}}=7^{-\frac{5}{6}+{-\frac{7}{6}}}=7^{-\frac{12}{6}}=7^{-2}$

Next, the rule with converting negative exponents into positive ones is $x^{-m}=\frac{1}{x^m}$ . Apply this rule here:

$7^{-2}=\frac{1}{7^2}=\frac{1}{49}$

<u>Your final answer is 1/49.</u>

So an additional rule when it comes to exponents is $x^{\frac{m}{n}}=\sqrt[n]{x^m}$

In this case, your fractional exponent, x^9/7, would be converted to $\sqrt[7]{x^9}$ . However, I had just realized you can further expand this.

Remember the rule I had mentioned earlier about multiplying exponents of the same base? Well, you can apply it here:

$\sqrt[7]{x^9}=\sqrt[7]{x^7*x^2}=x\sqrt[7]{x^2}$

Your final answer would be $x\sqrt[7]{x^2}$

3 0

3 years ago

Alexeev081 [22]

This is not a question...please rephrase.

3 0

3 years ago

Kitty [74]

Hello,

$x^2-2x-27=-5x+1\\

x^2-2x+5x-27-1=0\\

x^2+3x-28=0\\

x^2+7x-4x-28=0\\

x(x+7)-4(x+7)=0\\

(x+7)(x-4)=0\\

x=-7\ or\ x=4\\
$

8 0

3 years ago