<u>Solution:</u> 48cm/s^2
<u>Working:</u>
A = s^2
Derivate s^2 (as it is area formula) which gives
dA/dt = 2s dL/dt
dL/dt = 6cm/s
Hence,
2(6) = 12
Side: 4cm
Hence,
dA/dt = (4)(12) = 48cm/s^2
<em>Feel free to mark this as brainliest :D</em>
<span>A. 54°; acute is the answer</span>
Answer:
Median: 11
Mean: 12
Mode: none
Step-by-step explanation:
M A T H W A Y helps
Mode is something that is repeated multiple times
so
Answer:
x=6, y=75
Step-by-step explanation:
Since the diameter of the circle is a perpendicular bisector to the chord, x and 6 must be equal and therefore x=6. It also bisects the arc, meaning that y=75. Hope this helps!
8^2 /2+5(15-7)
=64/2+75-35
=32+40
=72
<span><span>3<span>(<span>5−9</span>)</span></span>+<span>4<span>(<span>4−9</span>)
</span></span></span><span>=<span><span><span>(3)</span><span>(<span>−4</span>)</span></span>+<span>4<span>(<span>4−9</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span>4<span>(<span>4−9</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span><span>(4)</span><span>(<span>−5</span>)
</span></span></span></span><span>=<span><span>−12</span>+<span>−20
</span></span></span><span>=<span>−32
</span></span><span><span>10<span>(<span>9−18</span>)</span></span>−<span>32
</span></span><span>=<span><span><span>(10)</span><span>(<span>−9</span>)</span></span>−<span>32
</span></span></span><span>=<span><span>−90</span>−<span>32
</span></span></span><span>=<span><span>−90</span>−9
</span></span><span>=<span>−<span>99
</span></span></span><span><span>−<span>12<span>(<span>5−7</span>)</span></span></span>−<span>10<span>(<span>2−5</span>)
</span></span></span><span>=<span><span><span>(<span>−12</span>)</span><span>(<span>−2</span>)</span></span>−<span>10<span>(<span>2−5</span>)
</span></span></span></span><span>=<span>24−<span>10<span>(<span>2−5</span>)
</span></span></span></span><span>=<span>24−<span><span>(10)</span><span>(<span>−3</span>)
</span></span></span></span><span>=<span>24−<span>(<span>−30</span>)
</span></span></span><span>=<span>54</span></span>
