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10 months ago

Find the area of a triangle whose sides are 3cm,6cm and 7cm

Mathematics

1 answer:

nadya68 [22]10 months ago

4 0

Answer:

See below

Step-by-step explanation:

It is NOT a RIGHT triangle so you cannot use 1/2 b * h

but you can use Heron's Formula

Area = sqrt( s (s-a)(s-b)(s-c) )

s = semi- perimeter = (3+6+7)/2 = 8

Area = sqrt ( 8 ( 8-3)(8-6)(8-7) ) = sqrt ( 80) = 8.9 cm^2

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Is the decimal for 4/3 a repeating decimal?explain

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Answer:

yes

Step-by-step explanation:

4 divided by 3 is 1.33333 repeating

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3 years ago

What is an angle that is adjacent to <2? (PLEASE)

STatiana [176]

Answer:

∠3 and ∠1 are correct

Step-by-step explanation:

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3 years ago

A horse can run at TopSpeed of 18 1/5 miles in 1/3 hour. What is the unit rate of the top running speed of a horse in miles per

sineoko [7]

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Answer:

54 3/5 mi/h

Step-by-step explanation:

To find miles per hour, divide miles by hours.

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8 0

2 years ago

find the zeros of following quadratic polynomial and verify the relationship between the zeros and the coefficient of the polyno

Cloud [144]

Answer:

$\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

$f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}$

To find the zeros, set the function to zero and solve for x using the quadratic formula.

$\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0$

Quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore,

a = 2√3
b = 5
c = - 4√3

Substituting the values into the quadratic formula:

$\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}$

$\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}$

$\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}$

$\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}$

$\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}$

$\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

The sum of the roots of a polynomial is -b/a:

$\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}$

The sum of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}$

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is: c/a

$\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2$

The product of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2$

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.

8 0

1 year ago