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3 years ago

4k/19m simplify this fraction

Mathematics

2 answers:

grin007 [14]3 years ago

7 0

4k/19m Final result : 4km /19

Step by step solution :Step 1 : k Simplify /19 Equation at the end of step 1 : k (4 • /) • m 19 Advertising SkipStep 2 :Final result : 4km/ 19

Olin [163]3 years ago

5 0

The answer is 4km over 19

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3 years ago

The data shows the speed, in miles per hour, of motorists on a stretch of road.

melomori [17]

Answer:

i. 18 miles per hour is an outlier

ii. the outlier decreases the mean speed

Step-by-step explanation:

An outlier in a given data is one of the values that is far greater or lesser compared to others. It affect the mean and standard deviation of a given data significantly.

From the given data, 18 is far too small compared to other values. This is certainly an outlier. This would affect the mean speed by decreasing the value.

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k/jkjln/kjln/j

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:

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

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

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

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

3 0

3 years ago

Need help? as soon as possible

Citrus2011 [14]

Answer:

$\sqrt{29}$ =5.385 (must look at the requirement of the test as how many decimal...)

Step-by-step explanation:

the squared is equal to 29 means area=29

square has four side equal.

Area=side * side

5 0

3 years ago