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patriots2833

patriots2833

29.07.2019 •

Mathematics

Factor completely: 5ab + 3ay + 5b + 3y (1 point) (5b + 3y)(a + 1) (5b - 3y)(a + 1) (5b + 3y)(a - 1) prime 5. factor completely: 4x3 + 28x2 + 7x + 49 (2 points) (x - 7)(4x2 + 7) (x - 7)(4x2 - 7) (x + 7)(4x2 + 7) (x + 7)(4x2 - 7) 6. factor completely: 21x3 + 35x2 + 9x + 15 (2 points) (3x - 5)(7x2 - 3) (3x - 5)(7x2 + 3) (3x + 5)(7x2 - 3) (3x + 5)(7x2 + 3) 7. factor completely: 10xy + 3y + 20ax + 6a (2 points) (10x - 3)(y - 2a) (10x - 3)(y + 2a) (10x + 3)(y + 2a) (10x + 3)(y - 2a)

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Ответ:

cathydaves

02.06.2020

4. ( 5b + 3y ) ( a + 1 )

5. ( x + 7 ) ( 4x² +7 )

6. ( 3x + 5 ) ( 7x² +3 )

7. ( 10x + 3 ) ( y + 2a )

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Ответ:

lolz55678

10.02.2022

The ball travel horizontally 37 feet before it hits the ground

Step-by-step explanation:

we have

A(x)=-0.02x^2+0.6x+5

where

x ----> is the horizontal distance in feet

A ---> is the altitude of the ball, in feet.

we know that

When the ball hit the ground, the altitude of the ball is equal to zero

so

For A(x)=0

-0.02x^2+0.6x+5=0

solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

-0.02x^2+0.6x+5=0

so

$a=-0.02\\b=0.6\\c=5$

substitute in the formula

$x=\frac{-0.6\pm\sqrt{0.6^{2}-4(-0.02)(5)}} {2(-0.02)}$

$x=\frac{-0.6\pm\sqrt{0.76}} {-0.04}$

$x=\frac{-0.6\pm0.87} {-0.04}$

$x=\frac{-0.6+0.87} {-0.04}=-6.75$

$x=\frac{-0.6-0.87} {-0.04}=36.75$

therefore

The ball travel horizontally 37 feet before it hits the ground

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