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yarga [219]
2 years ago
15

H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.

Mathematics
1 answer:
natima [27]2 years ago
4 0

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

<h3>Given information-</h3>

The given function is,

h(x)=x^2+1

k(x)=x-2

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

(h+k)(x)=h(x)+k(x)

(h+k)(x)=x^2+1+x-2

(h+k)(2)=2^2+1+2-2

(h+k)(x)=5

2)The value of the function (h-k)(3),

(h-k)(x)=h(x)-k(x)

(h-k)(x)=x^2+1-x+2

(h-k)(3)=3^2+1-3+2

(h-k)(x)=9

3) The value of the function 3h(2)+2k(3)

3h(x)+2k(x)=3x^2+3+2x-2\times 2

3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2

3h(2)+2k(3)=17

Hence the value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

Learn more about the quadratic equation here;

brainly.com/question/2263981

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Answer is B
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3 years ago
Bill has enough money to buy no more than 1/2 pound of cheese. how much cheese could he buy?
quester [9]
Let
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3 years ago
Line segment AB has endpoints A(1,8) and B(7,−4). What are the coordinates of the point located 5/6 of the way from A to B?
const2013 [10]

The coordinate of the point is (6,-2)

<h3>How to determine the coordinate of the point?</h3>

The given parameters are:

A = (1,8)

B = (7,-4)

The location of the point (i.e 5/6) means that the ratio is:

m :n = 5 : (6 - 5)

m : n = 5 : 1

The coordinate of the point is then calculated as:

(x,y) = \frac{1}{m + n}* (mx_2 + nx_1,my_2 + ny_1)

So, we have:

(x,y) = \frac{1}{5 + 1}* (5 * 7 + 1 * 1 , 5 * -4 + 1 * 8)

Evaluate

(x,y) = \frac{1}{6}* (36 , -12)

Evaluate the product

(x,y) = (6,-2)

Hence, the coordinate of the point is (6,-2)

Read more abut line segment ratio at:

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2 years ago
20 POINTS<br> How do you find the interquartile range
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Answer:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

(Khan Academy)

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3 years ago
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<img src="https://tex.z-dn.net/?f=Determine%20the%20unknown%20quantity.%20Show%20your%20work.8%201%2F2%20over%203%202%2F3%20%3D%
san4es73 [151]

um approximatly 0.34519202904

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2 years ago
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