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3 years ago

Please help asap, no rocky!

Mathematics

2 answers:

Naddika [18.5K]3 years ago

5 0

No, because it has a constant rate of change

damaskus [11]3 years ago

3 0

Answer:

first one and i need brainlist to level up

Step-by-step explanation:

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2 years ago

Plz plz help fast <br> I will mark Brainlyist

Licemer1 [7]

Answer:

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Step-by-step explanation:

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2 years ago

Plz help!:> also your so awesome, and I hope you have an amazing day!<br><br>

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3 years ago

A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$

Verify that $(9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark$

Therefore, the component form of this vector is $\vec{v}=\boxed{}\approx \boxed{}$

6 0

3 years ago

A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th

Temka [501]

The boat went 5hours upstream, let's say it has a "still water" speed rate of "b", it went 100miles... however, going upstream is going against the current, let's say the current has a speed rate of "c"

so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

now, recall your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
\qquad 50=\boxed{20+c}+c
\end{cases}$

solve for "c", to see what's the current's speed

what's "b"? well 20+c = b

4 0

3 years ago