Question

can you help me with my math questions

Answer 1

Answer:

1.  15/8   and -15/8

2. Yes it is a  perfect square.

3. a)  2.69 in

    b) 6 in

    c) 17.10 in

Step-by-step explanation:

1. Find the factors of 64f² - 225

64f² - 225 = 0  

(8f - 15)(8f +15) = 0

8f - 15 = 0      or        8f + 15 = 0

f = 15/8           or        f = -15/8

2. Find whether it is a perfect square p² + 14p + 49  

  p² + 14p + 49 = 0

  p² + 14p = -49

  p² + 14p + 7² = -49 + 7²

 (p + 7)² = 0

∴ p² + 14p + 49 =   (p + 7)²

Yes it is a  perfect square.

3 a)

36x² - 12x + 1 = 289

36x² - 12x = 288

x² - 1/3x + (1/6)² = 8 + 1/6

(x + 1/6)² = 49/6

x + 1/6 = ±2.8577

x = -3.02     and  x = 2.69

3. b)

25x² - 50x + 25 = 1225

x² - 2x + 1 = 49

x² -2x + 1 =48 + 1

(x + 1)² = 49

x = ±√49 - 1

x = 6  and  x = -8

3. c)

49x² - 56x + 16 = 289

49x² - 56x  = 289 - 16

49x² - 56x  = 273

x² - 1.1429x + 0.57² = 273 + 0.57²

(x - 0.57)² = 273.3265

x = ±√273.3265 + 0.57

x = 17.10    or x = -15.96

Answer 2

$p^2+2(7)p+7^2=(p+7)^2$

Justification:

Given:  $p^2+14p+49$

Split the middle term;

$=p^2+7p+7p+49$

Factor:

$=p(p+7)+7(p+7)$

$=(p+7)(p+7)$

$=(p+7)^2$

QUESTION 3a

The given expression for the area of the rectangle is  $36x^2-12x+1$.

This is equal to the indicated area which is $289\;in^2$

This implies that

$36x^2-12x+1=289$

$\Rightarrow (6x-1)^2=289$

$\Rightarrow (6x-1)(6x-1)=289$

$\Rightarrow l=(6x-1),w=(6x-1)$

This implies that, the dimensions of the rectangle are equal;

Using the square root method, we obtain

$\Rightarrow (6x-1)=\pm \sqrt{289}$

$\Rightarrow (6x-1)=\pm 17$

$\Rightarrow 6x=1\pm 17$

$\Rightarrow 6x=18 \:or\:6x=-16$

$\Rightarrow x=3 \:or\:x=-\frac{2}{3}$

We discard the negative value.

The side length of this rectangle is

$\Rightarrow l=(6(3)-1)=17,w=6(3)-1=17$

QUESTION 3b

The given expression for the area of the rectangle is  $25x^2-50x+25$.

This is equal to the indicated area which is $1225\;in^2$

This implies that

$25x^2-50x+25=1225$

$25(x^2-2x+1)=1225$

$(5(x-1))^2=35^2$

$5(x-1)5(x-1)=49$

$\Rightarrow l=5(x-1),w=5(x-1)$

This implies that, the dimensions of the rectangle are equal;

Using the square root method, we obtain

$\Rightarrow 5(x-1)=\pm \sqrt{1225}$

$\Rightarrow 5(x-1)=\pm 35$

$\Rightarrow x=1\pm 7$

$\Rightarrow x=8 \:or\:x=-6$

We discard the negative value.

The side length of this rectangle is

$\Rightarrow l=5(8-1)=35,w=5(8-1)=35$

QUESTION 3c

The given expression for the area of the rectangle is  $49x^2-56x+16$.

This is equal to the indicated area which is $289\;in^2$

This implies that

$49x^2-56x+16=289$

$(7x-4)^2=289$

$(7x-4)(7x-4)=289$

$\Rightarrow l=(7x-4),w=(7x-4)$

Applying the laws of indices gives;

$(7x-4)^2=17^2$

This implies that;

$7x-4=17$

$7x=21$

$x=3in.$

The side length of this rectangle is

$\Rightarrow l=7(3)-4=17\:in.,w=7(3)-4=17\:in.$

Dont forget that the square is also a rectangle.