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3 years ago

Classify each figure on the triangle

Mathematics

2 answers:

lara31 [8.8K]3 years ago

6 0

My answer video hope it helps you welp welp

ipn [44]3 years ago

5 0

All the triangles are isosceles meaning that they have to sides that are equal

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What is<br> 2x+3y=17<br> 3x+2y= -0.5

Flauer [41]

1.x=1 y=5

2.x=0 y=1/4

4 0

3 years ago

-2(v-2)=-3-2v plz help its home work

Svetach [21]

-2(v-2)=-3-2v doesn't have a solution

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3 years ago

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$

The maximum is 4

6 0

3 years ago

Knowing that 8<br> Solve in this format: <br> _

kifflom [539]

What? This isn’t a question

4 0

3 years ago

Can someone explain to me the meaning of x<-10

AlexFokin [52]

Answer: x<-10

Step-by-step explanation:

the < in the math equation means that the number to the right is greater than the number to the left. x can not be -10 it has to be lower (smaller).

3 0

4 years ago

Read 2 more answers