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AlladinOne [14]
1 year ago
5

Write the system of linear equations shown by the graph. Use slope-intercept form.

Mathematics
1 answer:
Andre45 [30]1 year ago
3 0

The required system of linear equations is y = -4x + 1 and y = 5x - 8.

Given that,
From the graph,
system of linear equation is to be determined.

Here,
The slope-intercept form of linear equation.
y = mx + c   - - - - - (1)


Equation of line 1
Since point (-1, 5), (0, 1) lies on the line 1.
The slope of line 1,
m = 1-5 / 0+1
m = -4
Now put this slope and 1 point in the equation.
5 = -4(-1) + c
c = 5 - 4
c = 1
now, put m and c in equation 1
y = - 4x + 1

Similarly, the equation of line 2 is given as y = 5x - 8.

Thus, the required system of linear equations is y = -4x + 1 and y = 5x - 8.

Learn more about lines here:

brainly.com/question/2696693

#SPJ1






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Evaluate S5 for 300 + 150 + 75 + … and select the correct answer below. 18.75 93.75 581.25 145.3125
Savatey [412]

we have that

300 + 150 + 75 +...

Let

a1=300\\ a2=150\\ a3=75

we know that

\frac{a2}{a1} =\frac{150}{300} \\\\ \frac{a2}{a1}=0.5 \\ \\ a2=a1*0.50

\frac{a3}{a2} =\frac{75}{150} \\\\ \frac{a3}{a2}=0.5 \\ \\ a3=a2*0.50

so

a(n+1)=an*0.50

Is a geometric sequence

Find the value of a4

a(4)=a3*0.50

a(4)=75*0.50

a(4)=37.5

Find the value of a5

a(5)=a4*0.50

a(5)=37.5*0.50

a(5)=18.75

Find S5

S5=a1+a2+a3+a4+a5\\ S5=300+150+75+37.5+18.75\\ S5=581.25

therefore

the answer is

581.25

Alternative Method

Applying the formula

S_n=\frac{a_1 (1-r^n)}{1-r} \\\\a_1=300 \\ r=\frac{1}{2}\\\\ S_5=\frac{300(1-(\frac{1}{2})^5)}{1-\frac{1}{2}}\\\\=\frac{300(1-\frac{1}{32})}{\frac{1}{2}}\\\\=\frac{300 \times \frac{31}{32}}{\frac{1}{2}}\\\\=\frac{75 \times \frac{31}{8}}{\frac{1}{2}}\\\\=\frac{\frac{2325}{8}}{\frac{1}{2}}\\\\=\frac{2325}{8} \times 2\\\\=\frac{2325}{4}\\\\=581 \frac{1}{4}\\\\=581.25

therefore

the answer is

581.25

6 0
3 years ago
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