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denis23 [38]
3 years ago
10

What is an adjacent angle

Mathematics
1 answer:
Ainat [17]3 years ago
8 0

Answer:

Adjacent angles are two angles that have a common side and a common vertex (corner point) but do not overlap in any way.For example, two pizza slices next to each other in the pizza box form a pair of adjacent angles when we trace their sides.

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A rectangle has an area of 24 square units and a side length of 2 3/4 units. Find the other side length of the rectangle. (Answe
Paul [167]
The answer is 8 8/11
4 0
3 years ago
What value(s) of x will make each equation below true?
kvasek [131]

Answer:

a X=0

b X=0

c X= undetermined

5 0
3 years ago
Read 2 more answers
Give an equation of a line that is parallel to −3x + 7y =4. Explain why your line is parallel to the given line and how you arri
koban [17]

Answer:

  3x -7y = 0

Step-by-step explanation:

Parallel lines have the same slope.

Changing the constant in a linear equation like this only changes the y-intercept. It has no effect on the slope of the line. So, we can change the constant from 4 to 0 and we will have a line with the same slope, parallel to the original, but with a different y-intercept.

The "standard form" of the equation of a line has the leading coefficient positive. We can make that be the case by using the multiplication property of equality, multiplying both sides of the equation by -1.

Parallel line:

  -3x +7y = 0

In standard form:

  3x -7y = 0

3 0
2 years ago
Please help asap 20 pts
My name is Ann [436]
C because 5 is (5,0)
8 0
3 years ago
Anyone can help me solve this equation using cross multiplying
Natali5045456 [20]

9514 1404 393

Answer:

  x = 1 or 5

Step-by-step explanation:

The notion of "cross-multiplying" is the idea that the numerator on the left is multiplied by the denominator on the right, and the numerator on the right is multiplied by the denominator on the left. This looks like ...

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\ \longrightarrow\ (x-1)(3x-1)=(7)(2x-2)

Then the solution proceeds by eliminating parentheses, and solving the resulting quadratic equation.

  3x^2-4x+1=14x-14\\\\3x^2-18x+15=0\qquad\text{subtract $14x-14$}\\\\x^2-6x+5=0 \qquad\text{divide by 3}\\\\(x-1)(x-5)=0\qquad\text{factor}\\\\x\in\{1,5\}

_____

<em>Comment on "cross multiply"</em>

Like a lot of instructions in Algebra courses, the idea of "cross multiply" describes <em>what the result looks like</em>. It doesn't adequately describe how you get there. The <em>one and only rule</em> in solving Algebra problems is "<em>whatever is done to one side of the equation must also be done to the other side of the equation</em>." If you multiply one side by one thing and the other side by a different thing, you are violating this rule.

What looks like "cross multiply" is really "<em>multiply by the product of the denominators</em> and cancel like terms from numerator and denominator." Here's what that looks like with the intermediate steps added.

  \displaystyle \frac{x-1}{7}=\frac{2x-2}{3x-1}\\\\\frac{x-1}{7}\times7(3x-1)=\frac{2x-2}{3x-1}\times7(3x-1)\\\\(x-1)(3x-1)=(2x-2)(7)\qquad\textit{looks like}\text{ cross multiply}

8 0
2 years ago
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