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Zarrin [17]
3 years ago
8

Solve the quadratic equations

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

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

p=1/4 twice

for quadratic equation 3

x^2-400=0

x^2=400

x=√400

x=20

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

n^2=25

n=√25

n=5

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John flips 10 coins and lays them out in a straight line.
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For every coin, two options are possible. The formula for calculting all possibilities is p^{t} where p is possibilities per time and t is times.

Therefore, there are 2^{10} which is 1024 combinations.

The formula for calculating probability is \frac{w}{t} where w is the ways to achieve the desired outcome, and t is the total outcomes possible.

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3 years ago
Can someone please help me with this i don't understand the way my teacher explained it​
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Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

Length of curve=s=32.66

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3 years ago
4km in9hr, then 24 km in
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Answer:a,d

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