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2 years ago

PLS HELP WILL MARK U BRAINLIEST!! NO FAKE ANSWERS

Mathematics

1 answer:

GREYUIT [131]2 years ago

6 0

Answer: A

Step-by-step explanation:

First, focus on the left side of the inequality. Make the two fraction have common denominators:

$\frac{2}{3} *\frac{5}{5} -\frac{x}{5} *\frac{3}{3}$

Multiplying fractions is simple. Multiply through and through like you normally would:

$\frac{10}{15} -\frac{3x}{15}$

Since they have the same denominator, they can be combined into one fraction:

$\frac{10-3x}{15}$

Multiply both sides of the inequality by 15:

$10-3x$

Subtract both sides by 10:

$-3x$

$x>-\frac{2}{3}$

Okay so, we know the answer must be greater than -2/3. This immediately rules out answer C. Answer B is also not the solution because zero is not a natural number. The smallest natural number that is greater than -2/3 is answer A.

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Simplify completely quantity x squared plus 4 x minus 45 all over x squared plus 10 x plus 9 and find the restrictions on the va

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Answer:

A) Quantity x minus 5 over quantity x plus 1, where x≠-1 and x≠-9

Step-by-step explanation:

$\frac{x^2 + 4x - 45}{x^2 + 10x + 9}$

Simplifying the numerator first:

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On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is: