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4 years ago

Plz help me get the right answer ASAP

Mathematics

2 answers:

Dmitry [639]4 years ago

5 0

X=-2 because 8÷-4 is -2

Pie4 years ago

5 0

Answer:

Step-by-step explanation:

$-4x=8\qquad\text{divide both sides by (-4)}\\\\\dfrac{-4x}{-4}=\dfrac{8}{-4}\\\\x=-2$

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Write down the value of the 6 in the number 2693

Goshia [24]

Answer:

6 = 600

Step-by-step explanation:

in a decimal number the digits to the left of the decimal point represent a whole number

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3 years ago

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Solve cos theta=- sqrt (3)/2 on [0, 2pi) Write the first solution or lowest on [0, 2pi) . Round to the nearest hundredth .

julsineya [31]

Answer:

2.62 radians

Step-by-step explanation:

The angle is found from its cosine using the inverse cosine function.

cos(θ) = (-√3)/2

θ = arccos((-√3)/2) = 5π/6 ≈ 2.62 . . . radians

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3 years ago

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notka56 [123]

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Mother nature

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3 years ago

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A canister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed t

solmaris [256]

Answer:

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

Step-by-step explanation:

A function<em> F</em> is called an antiderivative of <em>f</em> on an interval <em>I</em> if $F'(x) = f(x)$ for all x in <em>I.</em>

Recall that if the object has position function $s=f(t)$ , then the velocity function is $v(t)=s'(t)$ . This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is $a(t)=v'(t)$ , so the velocity function is an antiderivative of the acceleration.

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by $g$ . For motion close to the ground we may assume that $g$ is constant, its value being about $9.8 \:{\frac{m}{s^2}}$ .

We know that the acceleration due to gravity is given by

$a(t)=-9.8$

and the antiderivative is velocity

$v(t)=\int a(t)\,dt\\v(t)=\int -9.8\,dt\\v(t)=-9.8t +C$

We know that the canister was dropped, so the initial velocity at t = 0 is zero, this fact let us know the value of C.

$v(0)=9.8(0)+C\\C=0$

The antiderivative of velocity is the position

$s(t)=\int v(t) \, dt\\s(t)=\int -9.8t \, dt\\s(t)=-4.9t^2+C$

To find the value of the constant C, we know that the height was 500 m at t = 0, this means $s(0)=500$

$500=-4.9(0)^2+C\\C=500$

$s(t)=-4.9t^2+500$

Using the fact that at the time of impact the height s(t) is zero we can compute the total time of the fall:

$s(t)=-4.9t^2+500=0\\\\-4.9t^2=-500\\\\t^2=\frac{5000}{49}\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\t=\sqrt{\frac{5000}{49}},\:t=-\sqrt{\frac{5000}{49}}$