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r-ruslan [8.4K]
2 years ago
11

practice factoring a difference of squares. which expressions represent a perfect square monomial and its square root? check all

that apply. 121; 11 4x2; 2x 9x2 – 1; 3x - 1 25x; 5x 49x4; 7x2
Mathematics
1 answer:
ohaa [14]2 years ago
8 0

The perfect square monomial and its square root are shown in options 1, 2, and 5.

  • A perfect square in mathematics is an expression that factors into two equally valid expressions. A monomial is a single phrase that is made up of the product of positive integer powers of the constants, variables, and constants. Consequently, a monomial that factors into two monomials that are the same is called a perfect square monomial.
  • 1) 121, 11
  • 11² = 121
  • A perfect square monomial and its square root are represented by this equation.
  • 2) 4x², 2x
  • (2x)² = 4x²
  • A perfect square monomial and its square root are represented by this equation.
  • 3) 9x²-1, 3x-1
  • (3x-1)² = 9x²- 6x +1
  • This phrase does not depict a square monomial and its square root in perfect form.
  • 4) 25x, 5x
  • (5x)² = 25x²
  • This phrase does not depict a square monomial and its square root in perfect form.
  • 5) 49(x^4), 7x²
  • (7x²)² = 49(x^4)
  • A perfect square monomial and its square root are represented by this equation.

To learn more about monomial, visit :

brainly.com/question/9183135

#SPJ4

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Solve the equation for y.<br> 3x + jy= 7
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Answer:

y=\frac{7-3x}{j}

Step-by-step explanation:

First, subtract 3x. This leaves you with jy=7-3x. Then, divide by j, leaving you with y=\frac{7-3x}{j} .

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If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta​
Ivahew [28]

Step-by-step explanation:

<h3>Need to FinD :</h3>

  • We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

  • PQ = Opposite side
  • QR = Adjacent side
  • RP = Hypotenuse
  • ∠Q = 90°
  • ∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}

Since, we know that,

  • cosθ = 5/13
  • QR (Adjacent side) = 5
  • RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}

\rule{200}{3}

\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}

Where,

  • Opposite side = 12
  • Hypotenuse = 13

Therefore,

\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}

  • By substituting the values, we get,

\rule{200}{3}

\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the required answer is 17/7.

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3

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a5 = a* r^4 ( from general formular of a GP)

= 243 * (1/3)^4 =. 243 * 1/81 = 3

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