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Mkey [24]
3 years ago
14

Asafa runs 0.75 kilometers in 6 minutes. Jemaine runs 3,000 meters in half an hour. Who ran at a faster rate?

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

  • Asafa ran at a faster rate

Step-by-step explanation:

<u>Asafa's speed:</u>

  • 0.75 km/ 6 min = 750 m/ 6 min = 125 m/min

<u>Jemaine's speed:</u>

  • 3000 m / 30 min = 100 m/min

Asafa ran at faster rate as 125 m/min > 100 m/min

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What is the formula to <br>u-5=-4(w-1)
erastovalidia [21]

Answer: u = -4w + 9

<u>Step-by-step explanation:</u>

u - 5 = -4(w - 1)

u - 5 = -4w + 4

<u>   +5 </u>    <u>       +5  </u>

u         = -4w + 9

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3 years ago
If y=34 what is y50+y/60
zmey [24]
34 times 50 plus 34 divided by 60 equals 28.9
4 0
3 years ago
Read 2 more answers
) Evaluating a polynomial limit analytically You should have learned by now the process for finding the derivative of a polynomi
wolverine [178]

Answer:

This code or is program to find a given value of derivative of  a polynomial.

Step-by-step explanation:

We know already how to apply or make the procedures mathematically talking so this short program will eventually help you how to find logic.

// libraries

#include <stdio.h>

#include <conio.h>

//use to control floating elements

float poly(float a[], int, float);

//main

int main()

{

// Enter the degree of polynomial equation

float x, a[10], y1;

int deg, i;

printf("Enter the degree of polynomial equation: ");

scanf("%d", &deg);

printf("Ehter the value of x for which the equation is to be evaluated: ");

// Enter the coefficient of x to the power

scanf("%f", &x);

for(i=0; i<=deg; i++)

{

 printf("Enter the coefficient of x to the power %d: ",i);

 scanf("%f",&a[i]);

}

// The value of polynomial equation for the value of x

y1 = poly(a, deg, x);

 

printf("The value of polynomial equation for the value of x = %.2f is: %.2f",x,y1);

 

return 0;

}

/* function for finding the value of polynomial at some value of x */

float poly(float a[], int deg, float x)

{

float p;

int i;

 

p = a[deg];

 

for(i=deg;i>=1;i--)

{

 p = (a[i-1] + x*p);

}

 

return p;

}

8 0
3 years ago
Solve x^3-7x^2+7x+15​
ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0
3 years ago
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