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Rudik [331]
1 year ago
6

if a polygon layer and a line layer are intersected, the geometry of the output feature class may be chosen as a. Poinnts b. Lin

es c. Polygons
Mathematics
1 answer:
Sauron [17]1 year ago
8 0

if a polygon layer and a line layer are intersected, the geometry of the output feature class may be chosen as either a polygon or a line.

If the line layer is used to clip the polygon layer, the output feature class will be a polygon feature class that contains the part of the polygon layer that intersects with the line layer. The output will be a polygon feature class that contains the same attributes as the input polygon layer, plus a new attribute field indicating the line layer's object ID (OID) that was used to clip the feature. If the line layer is used to split the polygon layer, the output feature class will be a line feature class that contains the line features that were used to split the input polygon layer. The output will be a line feature class that contains the same attributes as the input line layer, plus a new attribute field indicating the polygon layer's OID that was used to split the features.

Learn more about polygon here

brainly.com/question/24464711

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3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
What is 1.75 kilometres as a fraction of 700 metres? <br>Give your answer in its simplest form​
aniked [119]

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Step-by-step explanation:

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5 0
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Ilia_Sergeevich [38]

Answer:

40

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That equals 140. Again, back to 180 degrees. We minus 180 degrees with 140 degrees which equals:

40 degrees.

4 0
3 years ago
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