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Xelga [282]
3 years ago
15

How to solve this problem

Mathematics
1 answer:
mart [117]3 years ago
7 0
Simply make a proportion. x is the long side for the small triangle and 4 is the short side. x is the short side of the medium triangle and 9 is the long side. Your proportion should look like this x/4=9/x. The cross multiply to get x squared = 36 . x=6 You can also draw a grid if you were taught to do so in school.
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Collins was 7 years 3 months old when he joined school. Today he is 13 years and 2 months. For how long has he been in school?
ivann1987 [24]

Answer:

He is about 6 years old (5 years 11 months)

Step-by-step explanation:

13 - 7 and 3 - 2

4 0
3 years ago
Which of the following are solutions to the equation below? Check all that apply. (3x - 5)2 = 19 ​
m_a_m_a [10]
What are the options to check off?
4 0
3 years ago
A hula hoop club bought new hula hoops. The state charges 6% sales tax. If the cost for one hula hoop was $17, what was the tota
True [87]
15.98 is the answer
8 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
What is the solution to the equation x+473 = 0
nikitadnepr [17]
-473 + 473=0 , so D.
5 0
3 years ago
Read 2 more answers
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