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3 years ago

A triangle has an area that is 125 m2. The measure of the base is 20 m so, h = ? m.

Mathematics

2 answers:

wolverine [178]3 years ago

5 0

Answer:

12.5

Step-by-step explanation:

12.5 multiplied by 20 is 250 which you divide by 2 to get the area. it plugs in to the triangle area formula which is 1/2 base times height.

Andreas93 [3]3 years ago

4 0

Answer:

Hello! answer: h = 12.5

Step-by-step explanation:

H = 12.5 because 12.5 x 20 = 250 then we divide by 2 and get our area 125. so h = 12.5

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There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the outp

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Answer:

<em> 15</em>
<em> 42,250</em>

Step-by-step explanation:

The given function is,

$P=40000+300x-10x^2$

where,

P = the total production of apples,

x = the number of trees added.

As the quadratic function has a negative leading coefficient, so it will open downward and at the vertex the value of function is maximum.

The vertex will be at $\left(-\dfrac{b}{2a},-f\left(\dfrac{b}{2a}\right )\right)$

The value of the function will be maximum at,

$x=-\dfrac{b}{2a}$

Putting the values,

$x=-\dfrac{300}{2\times (-10)}=\dfrac{300}{2\times 10}=\dfrac{300}{20}=15$

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Putting x=15 in f(x) will yield the maximum production of apples.

$P=40000+300(15)-10(15)^2=42,250$

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please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me

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Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH= $\sqrt{(5-0)^{2}+(1-1)^{2}}=5$ and

GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$

Now, mid point of JH= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

= $(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)$

Mid point of GI= $(\frac{5}{2},1)$

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

$(GI)^{2}=(IJ)^{2}+(JG)^{2}$ (1)

Now, JI= $\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20}$ and GJ= $\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}$

Substituting these values in (1), we get

$5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }$

$25=20+5$

$25=25$

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$ and HJ= $\sqrt{(0-5)^2+(1-1)^2}=5$ are equal, thus, GHIJ is a rectangle.

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