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soldier1979 [14.2K]

3 years ago

7

A high school as the department that 40 soccer uniforms at a cost of $3000 after soccer season they return some of the uniform i

s the only received $40 per uniform what was the difference between what they paid for each uniform and what they got for each return

1 answer:

love history [14]3 years ago

6 0

$3000 divided by 40 = 75...Mark as brainliest.......

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Which ordered pair represents the y-intercept of the graph of the equation y = -5x + 6?

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2 years ago

Help me! I need a better explanation!

jonny [76]

1) Those are polar coordinates

2) The expression inside cosine is the angle of the vector to be drawn

3) r is the length of the vector to be drawn.

For example:

- if you take angle =0 and r results to be =2, it means that you have to put a point over the x-axis at a distance of two units from the origin.

4) You have to look for notable values, like points on the axis and notable angles (0, 45°, 90°, 180 °,...)

5) Use derivation to find maximum and minimum values.

6) The graph of this function looks like the infinite symbol: ∞

7) Some notable points are: (0,0) ; (0,2) ; (0, - 2)

See attached the complete graph as a reference.

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3 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

2 years ago