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liberstina [14]
4 years ago
7

What is the standard form of the quadratic function that has a vertex at (3,-2) and goes through the point (4,0)?

Mathematics
1 answer:
nydimaria [60]4 years ago
5 0

Answer:

2x^2 - 12x + 16

Step-by-step explanation:

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Find the area of the figure.
34kurt

Hello from MrBillDoesMath!

Answer:

89 square meters

Discussion:

The area of the figure is the area of the full rectangle minus the area of the triangle .


Rectangle area = 13 * 8 = 104

Triangle area = 1/2 bh = 1/2 ( 13 - 4 -4) * 6 = 1/2(5)*6 = 15

So the area of the figure - 104 = 15 = 89


Regards,  

MrB


7 0
4 years ago
Use the information in the diagram to determine the height of the tree. The diagram is not a scale
dusya [7]
In the given diagram, we can note that the tree divides the hypotenuse of the triangle as well as the side where the tree is based into halves.

In a triangle, when a line is drawn joining between the midpoints of two sides, this line is always parallel to the third side and equals half its length.

Applying this concept to the given diagram here, we will find that the tree is joining the midpoints of two sides of the triangle. this means that the tree is parallel to the building and the height of the tree is half the height of the building.

Therefore:
height of tree = 0.5 * 120 = 60 ft
6 0
4 years ago
Solve the simultaneous equations <br>2p- 3q= 4<br>3p + 2q= 9​
Rzqust [24]

Answer:

q = 6/13, p = 35/13

Step-by-step explanation:

3p + 2q = 9

2p - 3q = 4

_________ (subtract the two equation from each other.)

p + 5q = 5

p = 5 - 5q

2(5 - 5q) - 3q = 4 (substitute value of p into equation)

10 - 10q - 3q = 4

10 - 13q = 4

-13q = -6

<u>q = 6/13</u>

3p + 2(6/13) = 9 (substitute value of q into equation)

3p + 12/13 = 9

3p = 105/13

<u>p = 35/13</u>

<u />

8 0
4 years ago
Rationalise:<br>(1)              4/(2+root3+root7)<br>(2)              4/(2root3+root5)
Temka [501]
\frac{4}{2+\sqrt3+\sqrt7}\cdot\frac{2-(\sqrt3+\sqrt7)}{2-(\sqrt3+\sqrt7)}=\frac{8-4\sqrt3-4\sqrt7}{2^2-(\sqrt3+\sqrt7)^2}=\frac{8-4\sqrt3-4\sqrt7}{4-3-2\sqrt{3\cdot7}-7}\\\\=\frac{8-4\sqrt3-4\sqrt7}{-6-2\sqrt{21}}=\frac{-2(2\sqrt3+2\sqrt7-4)}{-2(3+\sqrt{21})}=\frac{2\sqrt3+2\sqrt7-4}{3+\sqrt{21}}\cdot\frac{3-\sqrt{21}}{3-\sqrt{21}}\\\\=\frac{6\sqrt3-2\sqrt{63}+6\sqrt7-2\sqrt{147}-12+4\sqrt{21}}{3^2-(\sqrt{21})^2}=\frac{6\sqrt3-2\sqrt{9\cdot7}+6\sqrt7-2\sqrt{49\cdot3}-12+4\sqrt{21}}{9-21}

=\frac{6\sqrt3-6\sqrt7+6\sqrt7-14\sqrt3-12+4\sqrt{21}}{-12}=\frac{-8\sqrt3+4\sqrt{21}-12}{-12}=\frac{-4(2\sqrt3-\sqrt{21}+3)}{-12}\\\\=\frac{2\sqrt3-\sqrt{21}+3}{3}

==============================================================

\frac{4}{2\sqrt3+\sqrt5}\cdot\frac{2\sqrt3-\sqrt5}{2\sqrt3-\sqrt5}=\frac{8\sqrt3-4\sqrt5}{(2\sqrt3)^2-(\sqrt5)^2}=\frac{8\sqrt3-4\sqrt5}{4\cdot3-5}=\frac{8\sqrt3-4\sqrt5}{12-5}\\\\=\frac{8\sqrt3-4\sqrt5}{7}
4 0
4 years ago
Read 2 more answers
Please help I don't know what I'm doing wrong :/
vazorg [7]
You need to find the area of a kite in which case you would solve A=(pq)/2

P=16.3+16.3=32.6
Q=22.5+42.7
So pq=2,125.52
So A=1,062.76

So the answer is B
5 0
4 years ago
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