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lara [203]
3 years ago
5

Which fraction is equivalent to -7/8? A. 8/-7 B. 7/-8 C. -7/-8 D. 7/8

Mathematics
1 answer:
Galina-37 [17]3 years ago
7 0

Answer:

B. 7/-8

Step-by-step explanation:

it doesn't matter which number the negative sign is on, the result stays the same either way.

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Plz help, <br> Graph f(x)=|x−6|−4
sattari [20]

MrBillDoesMath!


Answer: I don't know how to draw graphs  on this website but here's the "picture".  Imagine the function y = absolute value(x). It looks like the English letter "V". The bottom of our V (i.e f(x) touches the x-axis when x = 6, so the "V" graph has been translated 6 units to the right of the origin. But  when x = 6 the value of f(x) is 0-4 = -4 so the tip of the V is located 4 units below the x axis. Summary:  g(x) looks like the absolute value function but  is translated 6 units to the right of the origin and 4 units down

MrB

3 0
3 years ago
Help please will mark brainliest!!
Anna11 [10]
20x20=400
pi(3.14) x 10^2(100)
100x3.14=314
400-100pi=85.8407 (then you can round if you need to)
8 0
3 years ago
Five tables and eight chairs cost $115; three tables and five chairs cost $70. Determine the cost of each table and of each chai
stiks02 [169]

$75

Step by step explanation:

4 0
3 years ago
What is the distance around this figure?
prisoha [69]

Answer:

40

Step-by-step explanation:

i think? i may have not understood the picture correctly, so if this is wrong i apologise

3 0
3 years ago
The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =
pentagon [3]

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du

so that

v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du

\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}

b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is s(0)=0\,\mathrm m. Then its displacement over the time interval [0, 3] is

s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}

c) The total distance traveled is the integral of the absolute value of the velocity function:

s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt

v(t) for 0\le t and v(t)\ge0 for \cos^{-1}\left(-\frac29\right)\le t\le3, so we split the integral into two as

\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt

=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt

\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}

4 0
3 years ago
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