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3 years ago

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I need help please

1 answer:

Triss [41]3 years ago

8 0

She paid $3.28 per gallon.

24.60 divided by 7.5 equals 3.28.

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How do I solve this inequality?

ch4aika [34]

The correct answer is: x ≥1

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3 years ago

HAPPY B DAY TOO meeeeeee

stich3 [128]

Answer:

happy birthday may god bless you

3 0

2 years ago

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Let a, b, c, and d be real numbers with a, c 6= 0. Prove that the lines y = ax+b and y = cx + d have the same x-intercept if and

Monica [59]

Step-by-step explanation:

We have got the lines :

$y=ax+b\\y=cx+d$

Both lines intercept the x-axis in the point :

$I = (i_{1} ,i_{2})$

In all point from x-axis the y-component is equal to 0.

$I=(i_{1},o)$

We replace the I point in the lines equations:

$0=a(i_{1})+b \\0=c(i_{1})+d$

From the first equation :

$0=a(i_{1})+b \\-b=a(i_{1})\\i_{1}=\frac{-b}{a}$

From the second equation :

$0=c(i_{1})+d\\ -d=c(i_{1})\\i_{1}=\frac{-d}{c}$

Then $i_{1}=i_{1}$

Finally :

$\frac{-b}{a}=\frac{-d}{c} \\\frac{b}{a}=\frac{d}{c} \\ad=bc$

y = ax + b and y = cx + d have the same x-intercept ⇔ad=bc

8 0

3 years ago

Which numbers represent the lower quartile, the median, and the upper quartile of the following set of numbers?

kotegsom [21]

Answer:

Lower quartile: 85; median: 97; upper quartile 115.5

Step-by-step explanation:

4 0

3 years ago

For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe

Rus_ich [418]

To solve the inequality $\frac{x^2+x+3}{2x^2+x-6}\ge \:0$

$\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:$

The numerator $x^2+x+3$ is not factorizable.

so factor the denominator $2x^2+x-6$ :

$2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\$

Now take $\mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)$

Then we get factor of the denominator as $\left(2x-3\right)\left(x+2\right)$

Thus $\frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}$

Now $\mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\$

Signs of $x^2+x+3>0$

$\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0$

$x\frac{3}{2}$ is the required solution of the given inequality.

5 0

3 years ago

Read 2 more answers