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3 years ago

476 Marks boole Calculate the area of this triangle. 17 cm 8 cm 15 cm

Mathematics

1 answer:

Naddika [18.5K]3 years ago

4 0

Answer:

Step-by-step explanation:

sides =8 cm, 17 cm and 15 cm

Area of triangle by Hero's Formula,

S=a+b+c/2

S=17+8+15/2

S=40/2

S=20

A= $\sqrt{s(s-a)(s-b)(s-c)}$

A= $\sqrt{20(20-17)(20-8)(20-15)}$

A= $\sqrt{(20)(3)(12)(5)}$

A= $\sqrt{3600}$

A=60 $cm^{2}$

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At what point on the parabola y = 3x^2 + 2x is the tangent line parallel to the line<br> y = 10x −2?

oksian1 [2.3K]

Answer:

( $\frac{4}{3}$ , 8 )

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 10x - 2 ← is in slope- intercept form

with slope m = 10

Parallel lines have equal slopes

then the tangent to the parabola with a slope of 10 is required.

the slope of the tangent at any point on the parabola is $\frac{dy}{dx}$

differentiate each term using the power rule

$\frac{d}{dx}$ (a $x^{n}$ ) = na $x^{n-1}$ , then

$\frac{dy}{dx}$ = 6x + 2

equating this to 10 gives

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x = $\frac{8}{6}$ = $\frac{4}{3}$

substitute this value into the equation of the parabola for corresponding y- coordinate.

y = 3( $\frac{4}{3}$ )² + 2

= (3 × $\frac{16}{9}$ ) + 2

= $\frac{16}{3}$ + $\frac{8}{3}$

= $\frac{24}{3}$

= 8

the point on the parabola with tangent parallel to y = 10x - 2 is ( $\frac{4}{3}$ , 8 )

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