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Bumek [7]
3 years ago
9

X+2=0 help thanks this is hard

Mathematics
2 answers:
djyliett [7]3 years ago
8 0
-2+2=0

I hope I helped
worty [1.4K]3 years ago
7 0
X + 2 = 0
<u>     -2   -2</u>
     x = -2
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Triangle A B C is shown. Angle A C B is a right angle. The length of the hypotenuse is 20.
kherson [118]

Answer:

The length of BC is needed because it is the side opposite ∠A.

Step-by-step explanation:

Given the right angles triangle as shown in the attachment, we can get sin(A) without using Pythagoras theorem. Instead we will use SOH CAH TOA trigonometry identity.

According to SOH:

Sin(A) = Opposite/Hypotenuse

Sin(A) = |BC|/|AB|

Opposite side of the triangle is the  side facing ∠A.

Based on the formula, we will need to get the opposite side of the triangle which is length BC for us to be able to determine sinA since the hypotenuse is given.

8 0
3 years ago
Read 2 more answers
Given the equation 5x − 4 = –2(3x + 2), solve for the variable. Explain each step and justify your process.
blagie [28]
A.
5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\&#10;5x-4=-2 \times 3x-2 \times 2 \\&#10;5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\&#10;11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\&#10;11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\&#10;x=0

B.
3(2x-4)=5x-1 \\&#10;6x-12=5x-1 \\&#10;\boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\&#10;11x=11 \\&#10;\boxed{x=11} \Leftarrow \hbox{the second mistake}

Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.

The correct solution:
3(2x-4)=5x-1 \ \ \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\&#10;3 \times 2x+3 \times (-4)=5x-1 \\ 6x-12=5x-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{subtract 5x from both sides} \\&#10;x-12=-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  |\hbox{add 12 to both sides} \\&#10;x=11
3 0
3 years ago
To determine whether a graph of a relation is also a function, Shayla declares that the y-axis is a vertical line and counts the
Mars2501 [29]

Answer:

No.

Step-by-step explanation:

The function is any equivalence relation in which there is only one output for every input. This means that the domain must be exhausted in order to gain all the elements of the range and each element of domain gets mapped to only one of the elements of the range. Vertical line test is used to determine whether a graph is a function or not. This test requires that vertical lines parallel to y-axis and including y-axis be drawn on the graph to see that each vertical line intersects the graph only once. This must be true for all the elements of the domain. Therefore, vertical line test requires to draw numerous vertical lines other than the y-axis since the definition of the function requires that. In short, mathematically, there will be a need for infinite number of vertical lines to perform the test. Therefore, Shayla is not applying the vertical line test correctly!!!

8 0
2 years ago
What is the range of the function represented by this graph? the graph of a quadratic function with a minimum value at the point
tatiyna

Answer:

see explanation

Step-by-step explanation:

The minimum is at the vertex (- 3, 1 ), that is y = 1

Since the graph is a minimum then it opens vertically up U

The range of the values of y are therefore

range [ 1, ∞ )

7 0
3 years ago
Find the equation of the exponential function represented by the table below:​
aivan3 [116]

Answer:

I believe its y=5(4)^x

Step-by-step explanation:

The equation is supposed to look like y=ab^x

a is basically where it all starts, so where the y meets the 0

- so the 5 in the y's place meets the 0

b is what you're multiplying by on the y's side, which is by 4 every time

and the x is the exponent and since there is no exponent you leave it as x

4 0
3 years ago
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