$\left\{\begin{array}{ccc}-5x+3y=3\\-4x+y=1&|\text{add 4x to both sides}\end{array}\right\\\\\left\{\begin{array}{ccc}-5x+3y=3\\y=4x+1&(*)\end{array}\right\\\\\text{Substitute }\ (*)\ \text{to the first equation}\\\\-5x+3(4x+1)=3\qquad\text{use distributive property}\\\\-5x+(3)(4x)+(3)(1)=3\\\\-5x+12x+3=3\qquad\text{subtract 3 from both sides}\\\\7x=0\qquad\text{divide both sides by (-5)}\\\\\boxed{x=0}\\\\\text{Substitute the value of x to}\ (*):\\\\y=4(0)+1\\y=0+1\\\boxed{y=1}\\\\Answer:\ x=0\ and\ y=1.$

5 0

2 years ago

Please help me with these 2 questions!!

IceJOKER [234]

Answer: 53. B similar

Step-by-step explanation:

8 0

2 years ago

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Simplify: 7j-6k-5j+4k also what is 30% of 520

alukav5142 [94]

The first one simplified is 2j-2k 30% of 520 is 156

7 0

3 years ago

Help ASAP. Thank you! Find the area of the shaded region. 80° 5 cm A=[?] cm2

pashok25 [27]

Answer:

Step-by-step explanation:

$\text{The area shaded outside of the triangle is}\\ \\ A=\frac{\pi r^2\theta}{360},\ r=5,\ \theta =80^o\\ \\ A=\frac{50\pi}{9}\\ \\ \text{Then we must add the area of the shaded triangle.}\\ \\ A=\frac{bh}{2}\\ \\ \frac{b}{2}=5sin40^o\\ \\ h=5cos40^o\\ \\ A=25sin40^o(cos40^o)\\ \\ \text{Total area is}\\ \\ A=\frac{50\pi }{9}+25sin40^o(cos40^o)\\ \\ A\approx 29.8cm^2$

8 0

2 years ago

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