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Anna11 [10]
3 years ago
5

What is 8.04=7+? can some one help me for this question please

Mathematics
2 answers:
Ivahew [28]3 years ago
6 0

Answer:

1.04

Step-by-step explanation:

8.04 - 7 = 1.04

I hope this helps!

navik [9.2K]3 years ago
5 0

Answer:

8.04=7+1.04is your answer.

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Darya [45]

Answer:

log(5)= 0.699

5 0
3 years ago
Read 2 more answers
H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.
natima [27]

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

4 0
3 years ago
What is the answer for y= 6x+8
goldenfox [79]

Answer:

x=-1/4 thats the answer alternative

6 0
3 years ago
Find the area of a regular hexagon with a said length of 10 centimeters and a apothem of 5 spuare root 3
erma4kov [3.2K]

Answer:

C

Step-by-step explanation:

the area (A) of the hexagon can be calculated as

A = \frac{1}{2} pa ( p is the perimeter and a the apothem )

p = 6 × 10 = 60 and a = 5\sqrt{3} , then

A = \frac{1}{2} × 60 × 5\sqrt{3}

   = 30 × 5\sqrt{3}

   = 150\sqrt{3}

   ≈ 259.8 cm² ( to the nearest tenth )

8 0
2 years ago
I need help ASAP!!!!!!
kipiarov [429]

Answer:

Not a function

Step-by-step explanation:

7 0
3 years ago
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