Two functions are shown. f(x) = 3x+1 g(x) =-3/2X-7 What is the value of x when f(x) = g(x)

Mathematics

1 answer:

katrin [286]3 years ago

8 0

Answer:

-1.778

Step-by-step explanation:

The two lines cross at the point (-1.778,-4.333), but the x value is -1.778

Hope this helps

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Anyone? please i need help

arsen [322]

Answer:
y = 5x - 7

Explanation:
The general form of the linear line in slope-intercept form is:
y = mx + c
where:
m is the slope
c is the y-intercept

1- getting the slope:
The slope of the line can be calculated using the following formula:
m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

We are given the points:
(3,8) representing (x₁,y₁)
(2,3) representing (x₂,y₂)
Substitute with the givens in the above equation to get the slope as follows:
m = $\frac{3-8}{2-3}$ = 5

Therefore, the equation now became:
y = 5x + c

2- getting the y-intercept:
To get the value of the c, we will use any of the given points, substitute in the equation we got in part 1 and solve for c. I will use point (2,3) as follows:
y = 5x + c
3 = 5(2) + c
3 = 10 + c
c = 3 - 10
c = -7

Based on the above, the equation of the line is:
y = 5x - 7

Hope this helps :)

6 0

2 years ago

Read 2 more answers

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

Heron's Formula

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,

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