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OLEGan [10]
3 years ago
9

What are the zeroes of y=x^2+14x+40

Mathematics
1 answer:
a_sh-v [17]3 years ago
3 0
y=x^2+14x+40 

y-x^2-14x-40=0 

In general, given a{x}^{2}+bx+c=0, there exists two solutions where

x=  \dfrac{14+2 \sqrt{y+9} }{-2} , \dfrac{14-2 \sqrt{y+9} }{-2} 

x=-7- \sqrt{y+9},-7+ \sqrt{y+9}
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Find the circumference of a circle with a diameter 1 1/2 inches
loris [4]

Answer:

C≅17.28

Step-by-step explanation:

Step one:If you divide 11 by 2 it would give you 5.5

Step two:Then use π which is 3.14

Step three:Multiply 5.5*3.14 which gives you 17.27 round to the nearest tenth would be 17.28

Hope it helps!

3 0
3 years ago
Kendra brought a quilt kit. In the kit were red, blue, and white cloth squares and the instructions below.
soldi70 [24.7K]

Blue=4

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I hope you are able to solve the question on your one shortly!

6 0
3 years ago
Plz help me plzzzzz asap
Varvara68 [4.7K]

Answer:

2 \frac{2}{9}

Step-by-step explanation:

First, we will turn 5 2/5 into an improper fraction:

5 * 5 = 25 + 2 = 27/5

So they worked for 27/5 days.

We want to divide the days worked by the distance to get the unit rate of distance per day:

\frac{12}{\frac{27}{5} } = \frac{12}{1} * \frac{5}{27}=\frac{60}{27}=\frac{20}{9}

So our answer is 2 \frac{2}{9}

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6 0
3 years ago
Solve the quadratic equations
Alexxx [7]

Answer:

for quadratic equation 1

y^2-10y+21=0

Step-by-step explanation:

y^2-7y-3y+21=0

y(y-7)-3(y-7)=0

(y-7)(y-3)=0

y-7=0,y-3=0

y=7,y=3

y=(3,7)

for quadratic equation 2

16p^2-8p+1=0

16p^2-4p-4p+1=0

4p(4p-1)-1(4p-1)=0

(4p-1)twice=0

4p-1=0,4p-1=0

4p=1

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for quadratic equation 3

x^2-400=0

x^2=400

x=√400

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for quadratic equation 4

-16m^2-8m-1=0

multiply the equation by -

16m^2+8m+1=0

16m^2+4m+4m+1=0

4m(4m+1)1(4m+1)=0

4m+1=0 twice

m=-1/4 twice

for quadratic equation 5

-3n^2+75=0

divide both side by -3

-3n^2/-3=-75/-3

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6 0
3 years ago
The line represented by the equation 2y - 3x = 4 has a slope of.....
S_A_V [24]
2y - 3x = 4 
2y = 3x + 4
y = 3/2*x + 2

y = kx + b ==>

k = 3/2 = 1,5
7 0
4 years ago
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