a carpenter goes through 2 3/5 boxes of nails finishing 3 1/2 rooves. How much would he use finishing 8 roves?

1 answer:

7 0

Find the unit rate first, and then multiply it by 8:

2 3/5 boxes
---------------- = 13/5 divided by 7/2.
3 1/2 roofs

You can obtain the correct result by inverting the divisor, 7/2, and then multiplying:

13/5 times 2/7. This comes out to 26/35. That represents 26/35 box of nails per roof.

To do 8 roofs (not roves), you'd need (26/35)(8/1) boxes: 5.94 boxes. Why not buy 6 boxes to be sure you'll have enough?

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0

3 years ago

Read 2 more answers

Can somebody please help me with #11 thank you

aniked [119]

The graph is not a function, as it does not pass the vertical line test. The lines at 2,-1 and 2,3 overlay and pass the vertical line more than once, meaning that the graph is not a function.

6 0

3 years ago

Answer to this math problem

Cerrena [4.2K]

Subsitute 1/3 for t

$18(\frac{1}{3})^2=18(\frac{1}{9})=\frac{18}{9}=2$