All categories

3 years ago

306 is 51% of what number?

Mathematics

2 answers:

Akimi4 [234]3 years ago

4 0

306 is 51% of the number 600

NeTakaya3 years ago

3 0

600 because is you put it in a proportion. With 51% over 100 and than 306 over a variable. you multiply 100 times 306 to get 30600. then you would divide it by 51 and you would get 600

You might be interested in

How do i do this? geometry

iren [92.7K]

Answer:

80?

Step-by-step explanation:

the total amount of degrees in a hexagon is 720. Divide that by 6 and you get 120. Each corner is 120. It doesnt have a right angle measure on it so it has to be lowerer then 90.

3 0

2 years ago

The original price is $60.The sales price is $45. What is the percent of discount?

MariettaO [177]

Hey there!

To start, first subtract 45 from 60 to find the amount of money taken off from the discount:
60-45
=15

Because 15 dollars have been taken off the original price, you can find the percent of discount by setting up this proportion where x represents the percent of discount:
15/60=x/100

Now, cross multiply and solve for x:
1500=60x
x=25

Therefore, your discount would be 25%.

Hope this helps and have a marvelous day! :)

5 0

3 years ago

1. The Royals softball team played 125 games and won 60 of them. What percent of the games did they lose? A. 48% B. 52% C. 65% D

Sauron [17]

Answer:

52% or B.

Step-by-step explanation:

3 0

3 years ago

When calculating the area of a trapezoid, how do you determine which sides are the bases?

ira [324]

Here is a video to help you with your question:

https://youtu.be/WaeD3qKc7Jc

3 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