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3 years ago

Please help I need this in ASAP

Mathematics

1 answer:

neonofarm [45]3 years ago

4 0

Answer:

Step-by-step explanation:

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74 is what percent of180

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3 years ago

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What is 40 over 30 in decimal form

Inessa05 [86]

Answer:

$\frac{40}{30}$ = 1.33333

Infinite 3's

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2 years ago

The length of each side of a square is 3 inches more than the length of each side of a small square. The sum of the areas of the

NeX [460]

Answer:

$\text{Smaller square's 15 inches, Bigger square's 18 inches}$

Step-by-step explanation:

GIVEN: The length of each side of a square is $3$ inches more than the length of each side of a small square. The sum of the areas of the square is $549$ inches.

TO FIND: the lengths of the sides of the two squares.

SOLUTION:

let the length of side of small square be $\text{x}$

Area of small square $=(\text{side})^2=\text{x}^2$

As length of each side of bigger square is $3\text{ inches}$ more than the smaller square

length of side of bigger square $=\text{x}+3\text{ inches}$

Area of bigger square $=(\text{x+3})^2$

Also

Sum of areas of both square $=549\text{ inch}^2$

$(\text{x})^2+(\text{x+3})^2=549$

$2\text{x}^2+6\text{x}+9=549$

$\text{x}^2+3\text{x}-270=0$

$\text{x}=15,-18$

as the length of side can never be negative

$\text{x}=15$

length of side of smaller square $=15\text{ inches}$

length of side of bigger square $=\text{x}+3\text{ inches}$ $=18\text{ inches}$

Hence the length of smaller and bigger square are $15\text{ inches}$ and $18\text{ inches}$ respectively.

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2 years ago

When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

bekas [8.4K]

a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ( $d_{tot}$ and the total time taken ( $t_{tot}$ ):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$ , we can substitute it into the previous equation to find the value of $t_2$ , the time the trucker drove in the city:

$v=\frac{200 mi}{3h +t_2}\\3h+t_2 = \frac{200 mi}{v}\\t_2 = \frac{200 mi}{v}-3h=\frac{200 mi}{57.14 mph}-3 h=0.50 h$

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3 years ago

Is the following statement always, never, or sometimes true?

pickupchik [31]

Sometimes...

2^-3 = 1/(2^3) = 1/8

-2^-3 = 1/(-2^3) = -1/8

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3 years ago

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