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puteri [66]
3 years ago
15

Gabriela has dinner at a cafe and the cost of her meal is

Mathematics
2 answers:
Maksim231197 [3]3 years ago
7 0

Answer: 51.75

Step-by-step explanation: 15 precent of 45 is 6.75

                                              Then add 45+6.75

ipn [44]3 years ago
4 0

Answer:

Her total including the bill will be $51.75

Step-by-step explanation:

15% of $45

= $6.75

$45 + $6.75 = $51.75

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The graph of an equation drawn through which two points would best represent the relationship between the number of miles and th
andreyandreev [35.5K]

Answer:

250

Step-by-step explanation:

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5 0
3 years ago
Pls help Im confused.​
Oksana_A [137]
The answer is x=13 because the equations equal each other
4 0
2 years ago
Read 2 more answers
The fountain is made up of two semicircles and a quarter circle. Find the perimeter and the area of the fountain. Round the peri
Andrej [43]
FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE.
 For this case what we have is the following:
 For the two semicircles we can model it as a complete circle.
 We have to then:
 
 Perimeter: 
 P = 2 * pi * r
 or
 P = pi * d
 Where,
 r = radius
 d = diameter
 Therefore the perimeter is:
 P = 10 * pi
 For the largest circle we have:
 radius = 10
 Perimeter:
 P '= 2pi10
 P '= 20pi
 1/4 since 1/4 circle:
 P '' = 20pi / 4 = 5pi
 Then, the total perimeter of the source is:
 Pt = P + P '' = 10pi + 5pi = 15pi
 Pt = 15 * (3.141592)
 Pt = 47.1239
 round
 Pt = 47.1 ft

 Area:
 The total area will be:
 A = A (two semicircles) + A (quarter big circle)
 A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
 A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
 A = 98.17477042 feet ^ 2
 Round:
 A = 98.2 feet ^ 2

 Answer: 
 Perimeter of the source: 
 Pt = 47.1 ft 
 Area of the source: 
 A = 98.2 feet ^ 2

5 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}

3 0
3 years ago
Please help with this question
goldfiish [28.3K]
Second answer box, try that
3 0
3 years ago
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