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Art [367]
3 years ago
5

Please Solve so i can complete!!!!!!!! Quick Fasts

Mathematics
1 answer:
Katen [24]3 years ago
3 0
I tried but couldn’t figure out sorry.
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Which of the following could be a function with zeros of —3 and 2? А f(x) = (x - 3)(x + 2) B f(x) = (x = 3)(x - 2) © f(x) = (x+3
Zinaida [17]

Answer:

C

Step-by-step explanation:

to find the zeros you look at what value you would fill in for x to make zero, so (x+3) would have a zero of -3 and (x-2) would have a zero of 2

6 0
3 years ago
Truth or false every fraction has a limited number of equivalent frections .
xeze [42]

Answer:

False.

Since the number of numbers in the world is infinite, there can be an infinite number of equivalent fractions.

4 0
3 years ago
Read 2 more answers
The equation of the line passing through the points (3, 16) and (5, 10) can be expressed as y = mx + b. Give the value of b
Alenkinab [10]

we are given two points as

(3, 16) and (5, 10)

firstly , we will find slope

slope is m

so, we can use formula

m=\frac{y_2-y_1}{x_2-x_1}

now, we can plug points

m=\frac{10-16}{5-3}

m=\frac{-6}{2}

m=-3

we can plug it in y=mx+b

we get

y=-3x+b

now, we can select any one points

and find b

(3,16)

x=3 and y=16

16=-3*3+b

we can find b

b=25

now, we can plug back b

and we get

y=-3x+25

so, equation is

y=-3x+25

and

b=25...............Answer

8 0
3 years ago
Find the equation of the line that contains the given point and the given slope. Write the equation in slope-intercept form.
stealth61 [152]

The slope-point form of a line:

y-y_0=m(x-x_0)

The slope-intercept form of a line:

y=mx+b

1.

m=6,\ (4,\ 1)\to x_0=4,\ y_0=1

Substitute

y-1=6(x-4)\qquad|\text{use distributive property}\\\\y-1=6x-24\qquad|\text{add 1 to both sides}\\\\\boxed{y=6x-23}

2.

m=-5,\ (6,\ -3)

Substitute

y-(-3)=-5(x-6)\qquad|\text{use distributive property}\\\\y+3=-5x+30\qquad|\text{subtract 5 from both sides}\\\\\boxed{y=-5x+24}

3.

m=-\dfrac{1}{2},\ (-8,\ 2)\\\\y-2=-\dfrac{1}{2}(x-(-8))\\\\y-2=-\dfrac{1}{2}(x+8)\\\\y-2=-\dfrac{1}{2}x-4\qquad|\text{add 2 to both sides}\\\\\boxed{y=-\dfrac{1}{2}x-2}

4.

m=0,\ (-7,\ -1)\\\\y-(-1)=0(x-(-7))\\\\y+1=0\qquad|\text{subtract 1 from both sides}\\\\\boxed{y=-1}

3 0
3 years ago
_________________<br>√(-√2-(-√2))²+(√3-2√3)²
xz_007 [3.2K]
_________________
√(-√2-(-√2))²+(√3-2√3)²= 3
7 0
3 years ago
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