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monitta
3 years ago
7

Kelly's made a painting in art class. The area of Kelly's paintings is 1.26 square feet. The height of her painting is 0.84. Wha

t is the length of Kelly's painting?
Mathematics
2 answers:
liq [111]3 years ago
5 0
The awnser would be i think 2.10
tia_tia [17]3 years ago
4 0
The answer is 2.20 for your questions you just need to add ok
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mylen [45]

Answer:

thanskkkkss broo!!

Step-by-step explanation:

5 0
3 years ago
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Write the equation of the line that has a slope of -3/2 and passes through (4, -3)
GREYUIT [131]

Answer:

y = -3/2x + 3

Step-by-step explanation:

y = -3/2x + b

-3 = -3/2(4) + b

-3 = -6 + b

3 = b

8 0
2 years ago
2x2 + 3x – 10 = 0<br> Find the discriminant for the quadratic equation
bija089 [108]

Answer:

The solutions are:

x=2

x=−5

Explanation:

2x2+3x−10=0

hope this helps (:

Step-by-step explanation:

4 0
2 years ago
Which function is a quadratic function? u(x) = –x 3x2 – 8 v(x) = 2x2 8x3 9x y(x) = x2 3x5 4 z(x) = 7x2 2x3 – 3.
Komok [63]

The function u(x)=-x+3x^{2} -8 is a quadratic function due to the highest exponent being 2.

Given functions are:

u(x)=-x+3x^{2} -8

v(x)=2x^{2} +8x^3+9x

y(x)=x^{2} +3x^5+4

z(x) =7x^{2} +2x^3-3

<h3>What is a quadratic function?</h3>

A function of the form f(x)=ax^{2} +bx+c with a\neq 0 is called a quadratic function means the highest exponent of polynomial should be 2.

Let us check one by one.

u(x)=-x+3x^{2} -8

Highest exponent =2

The function is also of form f(x)=ax^{2} +bx+c so u(x) is a quadratic function.

The functions v(x), y(x), and z(x) are having highest exponents as 3,5, and 3 respectively so they are not quadratic functions.

Hence, the function u(x)=-x+3x^{2} -8 is a quadratic function due to the highest exponent being 2.

To get more about quadratic functions visit:

brainly.com/question/1214333

7 0
2 years ago
Solve the system algebraically. Check your work.
Luden [163]

answer.

Answer:

x=2 and y=0 is the required result.

Step-by-step explanation:

We have been given system of equations:

5x+2y=105x+2y=10     (1)

And 3x+2y=63x+2y=6      (2)

We will use elimination method:

Multiply 1st equation by 3 and 2nd equation by 5 we get:

15x+6y=3015x+6y=30      (3)

15x+10y=3015x+10y=30      (4)

Now subtract  (4) from (3) we get:

-4y=0−4y=0

y=0y=0

Now, put y=0 in (1) equation:

5x+2(0)=105x+2(0)=10

5x=105x=10

x=2x=2

Hence, x=2 and y=0

8 0
3 years ago
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