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romanna [79]
3 years ago
15

A 10-foot board is to be cut into 3 pieces. Two of the pieces will be the same length and one piece will be 2 feet longer than t

he other two.
Which equation could be used to solve for the lengths?

A.) x + x + 2x = 10
B.) x + x + x - 2 = 10
C.) x + x + x = 10 + 2
D.) x + x + x + 2 = 10
Mathematics
2 answers:
aleksklad [387]3 years ago
8 0
2x+(x+2)=10  which could be expressed as:

x+x+x+2=10
garik1379 [7]3 years ago
7 0

Answer: D.) x + x + x + 2 = 10

Step-by-step explanation:

Given : A 10-foot board is to be cut into 3 pieces.

Two of the pieces will be the same length and one piece will be 2 feet longer than the other two.

Let x be the length of the each of the two equal pieces .

Then, length of third piece = x+2

According to the question , we have

Combined length of pieces = 10 foot

i.e. x+x+(x+2) = 10

i.e. x+x+x+2= 10  [on removing brackets]

Hence, the equation could be used to solve for the lengths : x + x + x + 2 = 10

So , the correct option is D.

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Answer:

13. c,

14. a.

Step-by-step explanation:

13.  The length of a side PQ with coordinates of

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       is: \sqrt{(x-w)^{2} +(y-z)^{2} }

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now, the easiest one will be when the vertices are on the coordinate axes.

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7 0
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}

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