Answer:
The correct option is B.
Step-by-step explanation:
The given vertices are (-4,2),(1,2),(1,-3) and (-4,-3).
Plot these point on a coordinate plate. From the graph it is noticed that the given quadrilateral is a square.
Distance formula:

Use distance formula to find the side length.


Since both consecutive sides are equal therefore it is a square.
Area of a square is

Where, a is side length.
The side length of the square is 5. So, area of ABCD is


Therefore the area of quadrilateral is 25 units square. Option B is correct.
Let f=number of field goals and a=number of points after.
a+f=38
a+3f=70
.
a+3f=70
a+ f=38
-------subtract
2f=32
.
f=16 field goals he kicked last season.
This expression can only be combined.
Add <span>p^2</span><span> and </span><span>−<span>32/p</span></span><span> to get </span><span><span><span>(<span>p^2</span>p−1⋅32)/</span>p</span>.
</span><span>−4p+<span><span><span>p^2 </span>p−1⋅32/</span>p</span>+4
</span>Multiply <span>−1</span><span> by </span>32<span> to get </span><span><span>−32</span>.
</span><span>−4p+<span><span><span>p^2 </span>p−32/</span>p</span>+4
</span>Find the common denominator<span>.
</span><span><span><span>−4pp/</span>p</span>+<span><span><span>p^2 </span>p−32/</span>p</span>+<span><span>4p/</span>p
</span></span>Combine<span> into one </span>fraction<span>.
</span><span><span>−4pp+<span>p^2 </span>p−32+4p/</span>p
</span>Simplify the numerator.
<span><span>−4<span>p^2</span>+<span>p^3</span>−32+4p/</span>p
</span>Simplify the expression<span>.
</span>
<span><span><span>p^3</span>−4<span>p^2</span>+4p−32/</span>p</span>
Answer:
actually I am sorry I dont rly know but I need some points
Step-by-step explanation:
This is the missing equation that models the hieght and is misssing in the question:
<span>h= 7cos(π/3 t)
</span>
Answers:
<span>a. Solve the equation for t.
</span>
<span>1) Start: h= 7cos(π/3 t)
</span>
2) Divide by 7: (h/7) = <span>cos(π/3 t)
</span>
3) Inverse function: arc cos (h/7) = π/3 t
4) t = 3 arccos(h/7) / π ← answer of part (a)
b. Find the times at which the weight is first at a height of 1 cm, of 3
cm, and of 5 cm above the rest position. Round your answers to the
nearest hundredth.
<span>1) h = 1 cm ⇒ t = 3 arccos(1/7) / π</span>
t = 1.36 s← answer
2) h = 3 cm ⇒ t = 3arccos (3/7) / π = 1.08s← answer
3) h = 5 cm ⇒ 3arccos (5/7) / π = 0.74 s← answer
c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time.
Use the periodicity property of the function.
The periodicity of <span>cos(π/3 t) is 6.
</span><span>
</span><span>
</span><span>So, the second times are:
</span><span>
</span><span>
</span><span>1) h = 1 cm, t = 6 + 0.45 s = 6.45 s ← answer
</span>
2) h = 3 cm ⇒ 6 + 1.08 s = 7.08 s← answer
3) h = 5 cm ⇒ t = 6 + 0.74 s = 6.74 s ← answer