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Varvara68 [4.7K]
4 years ago
13

In the figure below, ray was constructed starting from rays and . By using a compass D and G were marked equidistant from E on r

ays and . The compass was then used to locate a point F, distinct from E, so that F is equidistant from D and G. For all constructions defined by the above steps, the measures of ∠DEF and ∠GEF:
SAT
1 answer:
AURORKA [14]4 years ago
6 0
If you draw line segments DF and FG, you can show ΔDEF ≅ ΔGEF by SSS (side-side-side congruence). So, ∠DEF ≅ ∠GEF because corresponding parts of congruent triangles are congruent. Have fun with.. life lol
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On monday, you have a single request: order a for 15,000 units. it must be fulfilled by a single factory. to which factory do yo
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The  factory do you send the order  is Seattle factory. Check more from the image attached.

<h3>What is the factory about?</h3>

The reason for my decision is Seattle factory since they have the lowest cost per unit.

Know that for one to make a decision as in the case above, one has to know the cost per unit for all the factory and then we can make comparison in terms of cost and pick the lowest one.

Therefore the factory of Wichita cost price is: $1

That of Seattle: $0.75

Omaha (y) 0.00001 (15000) + 0.97

=-0.15 + 0.97

= 0.82

Therefore, We will have to send the order to the Seattle factory because they have the lowest cost per unit.

Learn more about factory from

brainly.com/question/10543530

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Evaluate the line integral, where c is the given curve. C (x/y) ds, c: x = t3, y = t4, 1 ≤ t ≤ 4.
skelet666 [1.2K]

We have

x = t³   ===>   dx/dt = 3t²

y = t⁴   ===>   dy/dt = 4t³

Then with the given parameteriztion, the line integral along C of x/y is

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^3}{t^4} \sqrt{(3t^2)^2 + (4t^3)^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac1t \sqrt{9t^4 + 16t^6} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{\sqrt{t^4}}t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 \frac{t^2}t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \int_1^4 t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 32t \sqrt{9 + 16t^2} \, dt

\displaystyle \int_C \frac xy \, ds = \frac1{32} \int_1^4 \sqrt{9 + 16t^2} \, d\left(9+16t^2\right)

\displaystyle \int_C \frac xy \, ds = \frac1{32} \cdot \frac23 \left(9+16t^2\right)^{\frac32}\bigg|_1^4

\displaystyle \int_C \frac xy \, ds = \frac1{48} \left(265^{\frac32} - 25^{\frac32}\right) = \boxed{\frac{265\sqrt{265}-125}{48}}

6 0
3 years ago
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