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3 years ago

What are the x-intercept and y-intercept for this line: 5x − 3y = 15

Mathematics

1 answer:

Andrej [43]3 years ago

3 0

Answer:

5x + 3y = 15 are (3,0) and (0,5).

Step-by-step explanation:

To learn how to find the x intercept, we will change the y for 0 and solve for x.

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40 hours at 10.10 per hour = 404

10.10 x 1.5 = 15.15 overtime pay per hour

15.15 x 8 = 121.20

404 + 121.20 = $525.20 total

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What is the sum or difference of 517 37/50 + 312 3/100

Romashka [77]

The sum is 829.77, and the difference is 205.44

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4 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

A car is towed using a force of 1600N. The rope used to pull the car makes an angle of 25∘ with the horizontal. Find the work do

mezya [45]

Answer: 2900185 Joules

Step-by-step explanation:

Force used to tow the car is 1600N

Angle(α)= 25

Displacement = 2km

Recall that 1km= 1000m

Therefore 2km= 2000m

Workdone= FdCos α

W= 1600 * 2000 * cos 25

W=2900184.919

W= 2900185 Nm(when approximated to the nearest integer)

Since 1J = 1Nm

Work done = 2900185 J

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horsena [70]

Answer:

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Step-by-step explanation:

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3 years ago