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3 years ago

Sarah wins the couch race by 32 centimeters. How many millimeters is 32 centimeters

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

320

Step-by-step explanation:

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Write a linear equation that passes through the point (2,-9) and has a slope of -5

Doss [256]

Answer:

y=-5x+1

Step-by-step explanation:

y-y1=m(x-x1)

y-(-9)=-5(x-2)

y+9=-5x+10

y=-5x+10-9

y=-5x+1

3 0

4 years ago

A kite is stuck in a 36-ft tree. If the angle of elevation from the kite and the ground is 12, find the length of the kite.

yanalaym [24]

Answer:

173.15 ft.

Step-by-step explanation:

Please find the attachment.

Let L be the length of kite.

We have been given that a kite is stuck in 36-ft tree. The angle of elevation from the kite and the ground is 12.

We can see from our attachment that kite and tree form a right triangle. height of tree is opposite of angle of elevation and length of kite is hypotenuse.

Since we know that Sine relates the opposite and hypotenuse of a right triangle, so we will use sine to find the length of kite.

$sin(12)=\frac{36}{L}$

$L=\frac{36}{sin(12)}$

$L=\frac{36}{0.207911690818}$

$L=173.1504364105882792\approx 173.15$

Therefore, the length of kite is 173.15 ft.

4 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:

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

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

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

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