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vladimir2022 [97]

3 years ago

10

What do I have? what do I need ?what should I use?

1 answer:

Ulleksa [173]3 years ago

4 0

For a:
you have your opposite side
you need your hypotenuse
you should use the s.o.h

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Which number is irrational...please help!!

yKpoI14uk [10]

Answer:

D I think

Step-by-step explanation:

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3 years ago

Tpy6a [65]

Answer:

$187.74\:\mathrm{cm}$

Step-by-step explanation:

The area of a rectangle is given by $l\cdot w$ . Therefore, we can set up the following inequality:

$l\cdot 16 \leq 3,003.84$ .

Solving this inequality, we have:

$l \leq 187.74$ .

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3 years ago

All points having an x-coordinate of 1 whose distance from the point (-3,-6) is 5

vazorg [7]

Answer:

$They'll \: be \: on \: the \: circle:$

$(x+3)^2+(y+6)^2=25$

$x=1$

$(y+6)^2=9$

$y + 6 = 3$

$y = -3$

$(1,-3)$

$y + 6 = -3$

$y = -9$

$(1,-9)$

Step-by-step explanation:

$Thank \: you!!!!!!$

6 0

3 years ago

8 liters is the same as how many gallons? round anwer to the nearst tenth

Alexandra [31]

Answer:

3.785, but rounded to the nearest tenth is 3.8

Step-by-step explanation:

Hope this helps!

4 0

2 years ago

Read 2 more answers

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

4 0

3 years ago