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exis [7]
3 years ago
9

Plz help me plz I need help I will mark brainlest for the answer​

Mathematics
1 answer:
Alexxx [7]3 years ago
6 0

Answer:

2/3

Step-by-step explanation:

Simplifying the Complex Fraction

Convert Mixed Numbers to Fractions

56114=5654

Method 1 : LCD Multiplication

The LCD for 6 and 4 is 12

Multiply top and bottom by the LCD

12×5612×54=1015

convert to mixed numbers and

reduce fractions where possible

=1015=23

Method 2 : Fraction Division

Divide the top fraction by the bottom

(multiply top by reciprocal of bottom)

56÷54

=56×45

=2030

convert to mixed numbers and

reduce fractions where possible

=2030=23

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A circle passes through point (-2, -1) and its center is at (2, -1). Which equation represents the circle?
Anastaziya [24]

Answer:

  \text{A)}\qquad (x-2)^2+(y+1)^2=16

Step-by-step explanation:

The formula for a circle of radius r centered at (h, k) is ...

  (x -h)^2 +(y -k)^2 = r^2

Both of the given points are on the line y=-1. The distance between them is the difference of their x-coordinates, 2 -(-2) = 4. So, the radius of the circle is 4 and the equation becomes ...

  (x -2)^2 +(y -(-1))^2 = 4^2

  (x -2)^2 +(y +1)^2 = 16 . . . . . . . . . matches choice A

7 0
3 years ago
If f(x) = x2 - 2x and g(x) = 6x + 4 for which value of x does (t +g)(x)=0​
Vinil7 [7]

Answer:

-2

Assumption:

Find the value of x such that (f+g)(x)=0.

Step-by-step explanation:

(f+g)(x)=0

f(x)+g(x)=0

(x^2-2x)+(6x+4)=0

Combine like terms:

x^2+4x+4=0

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.

(x+2)(x+2)=0

(x+2)^2=0

So we need to solve:

x+2=0

Subtract 2 on both sides:

x=-2

Let's check:

(f+g)(-2)

f(-2)+g(-2)

((-2)^2-2(-2))+(6(-2)+4)

(4+4)+(-12+4)

(8)+(-8)

0

0 was the desired output of (f+g)(x).

7 0
3 years ago
Solve for x <br> (1 point)
Delicious77 [7]
It should be 13 since 13+8 = 21. 21-7 = 13
5 0
3 years ago
Select ALL the correct answers.<br> Select all functions that have a y-intercept of (0,5).
otez555 [7]
1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok

2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No

3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No

4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok

5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No

Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10 
7 0
3 years ago
How many times does the equation f(x)=2x^2+9x-5 intersect the x-axis
Komok [63]

Answer:

2

Step-by-step explanation:

f(x)=2x^2+9x-5

When we are find how many times it intersects the x axis, we are finding the zero's. Set the equation equal to zero

0=2x^2+9x-5

Factor the equation

0 = (2x+1) (x-5)

2*1

1*-5 = -5

2*-5 +1*1 = -9

This checks for the first last and middle terms so we factored correctly

Then using the zero product property

2x+1 = 0 and x-5 =0

2x = -1    x=5

x = -1/2  and x=5

This function crosses the x axis 2 times

8 0
3 years ago
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