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DIA [1.3K]
3 years ago
12

Solve for angle...............

Mathematics
1 answer:
puteri [66]3 years ago
3 0
I am pretty sure it is 108 at least that is what I got. Hope this helps
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A quadrilateral has the following coordinates:
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Step-by-step explanation:

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Which store is offering the best deal
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4 years ago
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Find the volume of revolution bounded by the curves y = 4 – x2 , y = x, and x = 0, and is revolved about the vertical axis.
aleksley [76]
4-x^2=x\\
x^2+x-4=0\\
\Delta=1^2-4\cdot1\cdot(-4)=1+16=17\\
x_1=\dfrac{-1-\sqrt{17}}{2}\\
x_2=\dfrac{-1+\sqrt{17}}{2}\\\\
\displaystyle
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(4-x^2-x)^2\,dx\\
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(16-4x^2-4x-4x^2+x^4+x^3-4x+x^3+x^2)\,dx\\
V=\pi\int\limits_0^{\dfrac{-1+\sqrt{17}}{2}}(x^4+2x^3-7x^2-8x+16)\,dx\\
V=\pi \left[\dfrac{x^5}{5}+\dfrac{x^4}{2}-\dfrac{7x^3}{3}-4x^2+16x\right]_0^{\dfrac{-1+\sqrt{17}}{2}}\\


The rest of solution in the attachment. 

There's a mistake in the picture
It shoud be
V=\pi\left(\dfrac{289\sqrt{17}-521}{60}\right)\approx35

7 0
3 years ago
Find the measure of b.<br> please help!
Assoli18 [71]
<h3>Answer:  40</h3>

=======================================================

Explanation:

The inscribed angle 20 degrees doubles to 2*20 = 40 which is the measure of the central angle, and the arc in which the inscribed angle subtends (or cuts off). This is due to the aptly named inscribed angle theorem.

------------

A slightly longer alternative path would be to do this:

The triangle with interior angles 20 and c is isosceles. Note how the missing angle up top is one of the congruent base angles, so the missing angle is 20 degrees. That means angle c is...

20+20+c = 180

40+c = 180

c = 180-40

c = 140

Then angle b is supplementary to this

b+c = 180

b+140 = 180

b = 180-140

b = 40

This path leads to the same answer. It's slightly longer, but it's a path you can take if you aren't familiar with the inscribed angle theorem.

In fact, this line of thinking is effectively how the inscribed angle theorem is proved as shown in the diagram below.

6 0
3 years ago
Find the product of 30−−√ and 610−−√. Express it in standard form (i.E., ab√).
Zina [86]

we are given

\sqrt{30} *\sqrt{610}

we can radical formula

\sqrt{a} *\sqrt{b}=\sqrt{a*b}

we get

\sqrt{30} *\sqrt{610}=\sqrt{30*610}

\sqrt{30} *\sqrt{610}=\sqrt{3*61*100}

we can  also write as

\sqrt{30} *\sqrt{610}=\sqrt{100}*\sqrt{3*61}

\sqrt{30} *\sqrt{610}=10\sqrt{183}............Answer


7 0
4 years ago
Read 2 more answers
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