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3 years ago

Can someone help me with this math homework please!

Mathematics

1 answer:

blondinia [14]3 years ago

4 0

Answer:

Step-by-step explanation:

y coordinates must be at the numerator, while x coordinates at the denominator

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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.

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Solve each system of equations by writing a new system that eliminates one of the variables.

Rufina [12.5K]

Answer:

$x=33\\y=-41$

Step-by-step explanation:

Let:

$2x+y=25\hspace{5}(1)\\and\\4x+3y=9\hspace{5}(2)$

We need to eliminate one of the variables, so let's use elimination method. First multiply (1) by 2

$2*(1)\equiv\hspace{5}4x+2y=50$

Now subtract (2) from 2*(1) in order to eliminate x:

$2*(1)-(2)\\4x-4x+2y-3y=50-9\\-y=41$

Solving for y:

Multiplying both sides by -1

$y=-41$

Finally, replacing the value of y in (1)

$2x-41=25$

Solving for x:

add 41 to both sides:

$2x=66$

Multiply both sides by 1/2:

$x=\frac{66}{2} =33$

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I need help on numbers 17&18.

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