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Musya8 [376]
3 years ago
5

The midsegment through the legs of an isosceles triangle is 4 in. Find the sides of the triangle if its perimeter is 20 in. 75 p

oints
Mathematics
2 answers:
tigry1 [53]3 years ago
8 0

mid-segment - mid-point on the two sides

smaller triangle w/ mid-segment is similar to the original triangle

the third side of original triangle=mid-segment *2/1

=8

two sides + third side=perimeter=20

two sides=20-8=12

each of the two equal sides=6"


Pani-rosa [81]3 years ago
4 0

The midsegment is half the length of the base. So the base = 2*4 = 8 in.

Since it is an isosceles triangle, the other two sides are equal in length.

Perimeter = sum of two equal sides + base = 20 in.

2x equal side + 8 = 20

equal side = 12/2 = 6

Each of the equal sides = 6"


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marysya [2.9K]

Answer:

x = 16.5

Step-by-step explanation:

0.5x + 1.75 = 10

Subtract 1.75 from both sides:

0.5x = 8.25

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Check your work!

0.5(16.5) + 1.75 = 10

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Answer:

\huge\boxed{p(x)=4x^3-20x^2+4x+300}

Step-by-step explanation:

\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has  a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2

p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)

\text{The y-intercept is at 300}.\\\\\text{For}\ w(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\\\\\text{y-intercept is}\ a_0\\\\\text{Therefore for}\ p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)\\\\\text{y-intercet is}\ 75\bigg(r(x)\bigg)\\\\75\bigg(r(x)\bigg)=300\qquad\text{divide both sides by 75}\\\\r(x)=4\\\\\text{Finally:}\\\\p(x)=(x^3-5x^2+x+75)(4)\qquad\text{use the distributive property}\\\\p(x)=(x^3)(4)+(-5x^2)(4)+(x)(4)+(75)(4)\\\\p(x)=4x^3-20x^2+4x+300

7 0
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