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timofeeve [1]
3 years ago
8

Pls help me

t="49 = {x}^{2} " align="absmiddle" class="latex-formula">
49=x2​
Mathematics
2 answers:
Nitella [24]3 years ago
7 0

Answer:

x=-7

Step-by-step explanation:

Jobisdone [24]3 years ago
4 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  {x}^{2} = 49

According to rule : x² = y, then x = ± √y

\qquad \sf  \dashrightarrow \:  x =   \pm\sqrt{49}

\qquad \sf  \dashrightarrow \: x=  \pm7

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5 0
4 years ago
What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?
Sati [7]

<u>Answer-</u>

<em>The coordinates of the orthocenter of △JKL is (-4, 8)</em>

<u>Solution-</u>

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are  J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117

As,

\Rightarrow 117=81+36

\Rightarrow JL^2=JK^2+KL^2

\Rightarrow \text{JKL is a right angle triangle}

\Rightarrow \angle K=90^{\circ}

Therefore, the vertex at K (-4, 8) is the orthocentre.

3 0
3 years ago
What is the simplified of this ?
Hatshy [7]

Answer:

The correct answer is A) x^6y^8

Step-by-step explanation:

The easiest way to find this is to first square the second term.

(-x^2y^3)^2 = (x^4y^6)

Now we multiply that by the first parenthesis.

(x^2y^2)(x^4y^6) = x^6y^8

4 0
3 years ago
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