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2 years ago

100-point Question!!!

Mathematics

1 answer:

FrozenT [24]2 years ago

4 0

Answer:

Two or more independent functions (say f(x) and g(x)) can be combined to generate a new function (say g(x)) using any of the following approach.

h(x) = f(x) + g(x)h(x)=f(x)+g(x) h(x) = f(x) - g(x)h(x)=f(x)−g(x)

h(x) = \frac{f(x)}{g(x)}h(x)=

g(x)

f(x)

h(x) = f(g(x))h(x)=f(g(x))

And many more.

The approach or formula to use depends on the question.

In this case, the combined function is:

f(x) = 75+ 10xf(x)=75+10x

The savings function is given as

s(x) = 85s(x)=85

The allowance function is given as:

a(x) = 10(x - 1)a(x)=10(x−1)

The new function that combined his savings and his allowances is calculated as:

f(x) = s(x) + a(x)f(x)=s(x)+a(x)

Substitute values for s(x) and a(x)

f(x) = 85 + 10(x - 1)f(x)=85+10(x−1)

Open bracket

f(x) = 85 + 10x - 10f(x)=85+10x−10

Collect like terms

mark as brainiest

f(x) = 85 - 10+ 10xf(x)=85−10+10x

f(x) = 75+ 10xf(x)=75+10x

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A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude

I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

( a ) = 35 cm

( b ) = 54 cm

( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

<u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

$\sf{ \pmb { \longrightarrow \: \sqrt{s(s - a)(s - b)(s - c)} }}$

where ( s ) :

$\sf \longrightarrow s = \dfrac{a + b + c}{2}$

Therefore, for the given triangle first we will calculate ( s )

$\begin {aligned}\quad & \quad \longmapsto \sf s = \dfrac{a + b + c}{2} \\ & \quad \longmapsto \sf s = \dfrac{35 + 54 + 61}{2} \\ & \quad \longmapsto \sf s = \dfrac{150}{2} \\ & \quad \longmapsto \sf s = 75cm \end{aligned}$

Now, Area of triangle will be:

$\begin{aligned}&:\implies \sf\quad \sf \: A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \: A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)} \\&:\implies \sf\quad \sf \: A = \sqrt{75 \times 40 \times 21 \times 14} \\ &:\implies \sf\quad \sf \: A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5} \\ &:\implies \sf\quad \sf \: A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5} \\ &:\implies \sf\quad \sf \: A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf A =939.15 {cm}^{2} }} \end{aligned}$

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

$\begin{aligned}&:\implies \sf\quad \sf \: Area =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times b \times h =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times 61 \times h = 939.15 \\&:\implies \sf\quad \sf \: h =939.15 \times \dfrac{2}{61} \\&:\implies \sf\quad \sf \: h = \dfrac{1818.3}{61} \\ &:\implies \sf\quad \boxed{ \pmb{\sf \: h =30.79 \: (approx)}} \end{aligned}$