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lakkis [162]
1 year ago
15

the length of the longer leg of a right triangle is 3 ft more than three times the length of the shorter leg. the length of the

hypotenuse is 4 ft more than three times the length of the shorter leg. find the side lengths of the triangle.

Mathematics
1 answer:
Ulleksa [173]1 year ago
5 0

with the pythagorean theorem

\begin{gathered} (4+3x)^2=x^2+(3+3x)^2 \\ 16+24x+9x^2=x^2+9+18x+9x^2 \\ 16+24x+9x^2=10x^2+18x+9 \\ 16+24x+9x^2-9=10x^2+18x+9-9 \\ 9x^2+24x+7=10x^2+18x \\ 9x^2+24x+7-18x=10x^2+18x-18x \\ 9x^2+6x+7=10x^2 \\ 9x^2+6x+7-10x^2=10x^2-10x^2 \\ -x^2+6x+7=0 \end{gathered}

using the formula of the quadratic equation

\begin{gathered} x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x1=\frac{-6+\sqrt{6^2-4\left(-1\right)\cdot\:7}}{2\left(-1\right)}=-1 \\ x2=\frac{-6-\sqrt{6^2-4\left(-1\right)\cdot\:7}}{2\left(-1\right)}=7 \end{gathered}

the length cannot be negative, therefore x=7

length of the shorter leg is: 7ft

length of the longer leg is: 3+3(7)= 24ft

length of the hypotenuse is: 4+3(7)= 25ft

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3 years ago
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3 years ago
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which of the following is a solution of 2cos^2x-cost-1=0 a. 0 degrees b. 150 degrees c. 180 degrees d. 210 degrees
Luba_88 [7]

Answer:

{0, 150} degrees

Step-by-step explanation:

Given 2cos^2x-cost-1=0, let's simplify this problem by temporarily replacing cos x with y:

2y^2 - y -1 = 0

This can be solved by factoring:  (2y + 1)(y - 1) = 0.  From this we get two solutions:  y = -1/2 and y = 1.

Remembering that we let y = cos x, we now solve:

cos x = -1/2 and cos x = 1.

Note that cos x = 1 when x = 0 and the "adjacent side" coincides with the hypotenuse.

cos x = -1/2 when the hypotenuse is 2 and the "adjacent side" is -1.  This has two solutions between 0 and 360 degrees:  150 degrees and  270 degrees.

Four answer choices are given.  Both (a) (0 degrees) and (b) (150 degrees) satisfy the original equation.  Thus, the solution set is {0, 150} (degrees).

3 0
3 years ago
Someone please help me
Tresset [83]
About 14.58% it’s 3499.2 14.59% ends up being around 3501.6
3 0
2 years ago
How much Zn and how much HCl should be used to produce 1.5 g ZnCl2
tatuchka [14]

Answer:

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Step-by-step explanation:

The required mass of reactants can be computed from the number of moles of product.

  moles of ZnCl₂ = (1.5 g)/(136.282 g) ≈ 0.0110066 moles

Then the mass of Zn required is ...

  mass of Zn = (0.0110066 mol)(65.382 g/mol) = 0.71963 g

Two moles of HCl are used in the reaction with each mole of Zn, so the mass of HCl required is ...

  mass of HCl = 2(0.0110066 mol)(36.458 g/mol) = 0.80256 g

_____

<em>Comment</em>

The discrepancy in the last two decimal places of the weights above versus the calculator output below comes from rounding of moles of Zn to 4 significant digits, instead of 5 or more. The number of moles of ZnCl₂ is closer to 0.0110066.

4 0
3 years ago
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