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3 years ago

Can you help me find each measure?

Mathematics

1 answer:

madam [21]3 years ago

4 0

Answer:

Step-by-step explanation:

1. the measure for HJ you can already see, start from h and draw a line to J. The measure shows 63 degrees.

2. start at F and draw a line to G first. the measure shows 65 degrees. Now continue the line from where you stopped at G, to H. There's no measure but you can see it is in a semicircle. A semicircle is 180 degrees andthe other two angles are 63 and 65.

So...

180=63+65+GH

subtract to get GH

180-63-65= 52

so the measure GH is 52 degrees. The full measure you are trying to find is FGH thought. So ad the 65 and the 52.

FGH =117 degrees.

3. The meaure is CDE. CD as you can see is a right angles, so 90 degrees. But there is no measure for DE. If you look to the angle vertical from DE which is BA. It measures 40 degrees. DE and BA are vertical so they are congruent. If DE equals 40 and CD equals 90, put them together and you get 130.

CDE= 130 degrees

4. Next measure is BCD. We already know CD is a 90 degree angle but BC is blank. You can see the measure BCD is in a semi circle. A semicircle equals 180, CD equals 90, and BA equals 40

so...

180= 90+40+BC

so subtract

180-90-40= 50

BC =50

so add BC=50 and CD=90. SO, BCD is 140 degrees.

5.The angle LMN is next. MN is 30 but LM is blank. LMN is in a semi circle.

A semicircle is 180 degrees.

so...

180=105+30+LM

subtract

180-105-30=45

LM = 45

Add LM=45 and MN=30

LMN is 75 degrees.

6.The last angle is LNP

If you look at it MP is a semi circle, so 180 degrees. And LM is 45 from our last question. so 180 +45=225

LNP=225

hope this helps

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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

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Let $P=(x,y,z)$ be an arbitrary point on the surface. The distance between $P$ and the given point $(7,11,0)$ is given by the function

$d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}$

Note that $f(x)$ and $f(x)^2$ attain their extrema, if they have any, at the same values of $x$ . This allows us to consider the modified distance function,

$d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2$

So now you're minimizing $d^*(x,y,z)$ subject to the constraint $z^2=xy$ . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

$\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)$

which has partial derivatives

$\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}$

Setting all four equation equal to 0, you find from the third equation that either $z=0$ or $\lambda=-1$ . In the first case, you arrive at a possible critical point of $(0,0,0)$ . In the second, plugging $\lambda=-1$ into the first two equations gives

$\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10$

and plugging these into the last equation gives

$z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5$

So you have three potential points to check: $(0,0,0)$ , $(2,10,2\sqrt5)$ , and $(2,10,-2\sqrt5)$ . Evaluating either distance function (I use $d^*$ ), you find that

$d^*(0,0,0)=170$
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$d^*(2,10,-2\sqrt5)=46$

So the two points on the surface $z^2=xy$ closest to the point $(7,11,0)$ are $(2,10,\pm2\sqrt5)$ .

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3 years ago

Can someone please help me find the measure of the exterior angle.

dimulka [17.4K]

Answer:

k = 140°

Step-by-step explanation:

The sum of the interior angles of any triangle = 180°.

64° + 76° + x = 180°

x = 40°

40° + k = 180° because they are supplementary angles

k = 140°

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Answer:

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Here's an example: Jim had a bag of 15 jelly beans. He eats 8. How many jelly beans does he have left? Take apart the number to make a 10. Then subtract.

Here, you need to "take apart" the 8 to make your 15 jelly beans 10, then you subtract the remainder of 8 to get your solution. In other words,

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On paper, the problem looks like this:

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