difference of squares means that both the terms are square terms. (also there must be a - symbol)
for example
y^2 - 4
square root of y^2 is y
square root of 4 is +2 as well as -2
so you would factorise it like this:
(y+2)(y-2)
1. y^4 has a square root of y^2 as y^2 × y^2 is y^4.
<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>-2 doesnt have a whole number square root so it is not a difference of squares.
2. 25 has a square root of 5. m^2 has a square root of m. n^4 has a square root of n^2. so this 25m^2n^4 is a square term.
1 has a square root of +1 and -1.
therefore, this one is a difference of squares. <u>(</u><u>5</u><u>m</u><u>n</u><u>^</u><u>2</u><u> </u><u>+</u><u>1</u><u>)</u><u> </u><u>(</u><u>5</u><u>mn^2</u><u> </u><u>-</u><u>1</u><u>)</u>
3. p^8 has a square root of p^4. q^4 has a square root of +q^2 and -q^2)
so it is a difference of squares. <u>(</u><u>p</u><u>^</u><u>4</u><u>+</u><u>q</u><u>^</u><u>2</u><u>)</u><u>(</u><u>p</u><u>^</u><u>4</u><u> </u><u>-</u><u>q</u><u>^</u><u>2</u><u>)</u>
4. 16x^2 is a square term as irs square root is 4x.
<em>h</em><em>o</em><em>w</em><em>e</em><em>v</em><em>e</em><em>r</em><em>,</em><em> </em>24 is not a square term.
therefore, it is not a difference of squares.
Answer:
$54.75
Step-by-step explanation:
0.05 x 365 = 18.35
18.25 x 3 = 54.75
Answer:
The correct options are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Given that RS is parallel to DC, we have;
∠BDC = ∠BRS (Angles on the same side of transversal)
Similarly;
∠BCD = ∠BSR (Angles on the same side of transversal)
∠CBD = ∠CBD = (Reflexive property)
Therefore;
ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency
2) Whereby ΔBCD ~ ΔBSR, we therefore have;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
Inverting both sides
BR/RD = BS/SC
3) From BR/RD = BS/SC the above we have by cross multiplication;
BR/RD = BS/SC gives;
BR × SC = RD × BR → (BR)(SC) = (RD)(BR).
