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Colt1911 [192]
3 years ago
12

Which of the following is NOT a way to prove triangles similar?

Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0
I think a because b and c and trying to explain the triangles
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Determine the intercepts of the line. 7x-5=4y-67x−5=4y−67, x, minus, 5, equals, 4, y, minus, 6 xxx-intercept: \Big((left parenth
sesenic [268]

Given:

The equation of line is

7x-5=4y-6

To find:

The x and y-intercept of the given line.

Solution:

We have,

7x-5=4y-6

Putting x=0, we get

7(0)-5=4y-6

Add 6 on both sides.

0-5+6=4y

1=4y

Divide both sides by 4.

\dfrac{1}{4}=y

So, the y-intercept is \left(0,\dfrac{1}{4}\right).

Putting y=0 in given equation, we get

7x-5=4(0)-6

Add 5 on both sides.

7x=-6+5

7x=-1

Divide both sides by 7.

x=\dfrac{-1}{7}

So, the x-intercept is \left(-\dfrac{1}{7},0\right).

4 0
3 years ago
Read 2 more answers
(Hehe...Please ignore the markings)
inysia [295]

Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

x+y=8, multiply each term by 4:

4*x=4x, 4*y=4y, 8*4=32

so, the equivalent system is: 4x+4y=32 and x-y=2

Solve the system of equations:

x-y=2 becomes x=y+2

plug into 4x+4y=32 to solve for y

4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

3 0
3 years ago
What is the initial value of the exponential function shown on the graph? 7 ts O 0 1 3 O2 2 4. 7 5 4 3 2 1 1 1 2 3​
Tamiku [17]

Answer:

D.(4)

Step-by-step explanation:

8 0
2 years ago
What is the solution set for the equation x2-5x +6 =0
Liula [17]
You want to get the x's on one side and all the other numbers on the other side of the equation.
2x-5x+6=0
2x-5x=-6
-3x=-6
x=2
5 0
3 years ago
Read 2 more answers
Could someone give me the answers do these please
Debora [2.8K]
5. (0,1)
6. (-2,-2)
7. (-1,-3)
8. (-3, 2)
5 0
3 years ago
Read 2 more answers
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