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3 years ago

Very urgent... I need it right now.. please help me with explanation!

Mathematics

1 answer:

aliina [53]3 years ago

4 0

Answer:

a) Water height, H(g) = 8g^2 + 3g -4 - [9g^2 -2g -5] = 8g^2 + 3g -4 -9g^2 + 2g +5 = -g^2 +5g +1

b) g = 1

H(g) = -(1)^2 + 5(1) + 1 = -1 + 5 + 1 = 5

g=2

H(g) = -(2)^2 + 5(2) + 1 = -4 + 10 + 1 = 7

g=3

H(g) = -(3)^2 + 5(3) + 1 = -9 +15 +1 = 5

c) Greatest height

Find the vertex of the parabole

The vertex is at the mid point between the two roots.

To find the roots you can use the quadratic equation

The result is g = 5/2 + [√29]/2 and g = 5/2 - [√29]/2

The middle poin is 5/2 = 2.5

Now find H(2.5) = -(2.5)^2 + 5(2.5) + 1 = 7.5 ≈ 7.3

hope this helps

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Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

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$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

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step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

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