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zvonat [6]
3 years ago
14

4. The range of a relation is given -3, 2, 0, 13, 17 Which domain makes the relation not a function? A. 0, 1, 2, 3, 4 B. 1,1, 2,

3,1 C. 5, 4, 2, 1,0​
Mathematics
1 answer:
faust18 [17]3 years ago
6 0
The domain that makes the relation not a function is c
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What is 7 to the 10 power
Over [174]
7 to the 10 power would be 70
5 0
3 years ago
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How would I solve for the parabolic equation with these two given points. Please answer ASAP!!
monitta
So

y=ax^2+bx+c
(x,y)
sub the points and solve

(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c

(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points

maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)

solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation

15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b

if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
7 0
3 years ago
Is anyone have the answer to 9,576,984.9874 x 98.84
algol [13]

Answer:

946589196.155

Step-by-step explanation:

Hope that helps

6 0
3 years ago
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Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.
Anna [14]

Answer:

a.\ m=2,\ \text{the line is rising}\\\\b.\ m=-\dfrac{5}{4},\ \text{the line is falling}\\\\c.\ m=0,\ \text{the line is horizontal}\\\\d.\ m\ is\ unde fined,\ \text{the line is vertical}

Step-by-step explanation:

The formula of a slope:

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

If

m > 0, then a line is rising

m < 0, then a line is falling

m = 0, then a line is horizontal

m is undefined, then a line is vertical

<h2>a.</h2>

(2, 1) and (4, 5)

m=\dfrac{5-1}{4-2}=\dfrac{4}{2}=2>0\to\text{rising}

<h2>b.</h2>

(-1, 0) and (3, -5)

m=\dfrac{-5-0}{3-(-1)}=\dfrac{-5}{4}=-\dfrac{5}{4}

<h2>c.</h2>

(2, 1) and (-3, 1)

m=\dfrac{1-1}{-3-2}=\dfrac{0}{-5}=0\to\text{horizontal}

<h2>d.</h2>

(-1, 2) and (-1, -5)

m=\dfrac{-5-2}{-1-(-1))}=\dfrac{-7}{0}\ \text{UNDEFINED}\to\text{vertical}

7 0
3 years ago
.........don't mind this I put this up by accident
JulijaS [17]

Answer:

Really?

Step-by-step explanation:

Ok....................

5 0
3 years ago
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