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marysya [2.9K]
3 years ago
8

11 + 2(z + 3) -z simplified

Mathematics
2 answers:
lbvjy [14]3 years ago
3 0

Answer: z + 17

Step-by-step explanation:

1) DISTRIBUTE:

11 + 2z + 6 - z

2) COMBINE LIKE TERMS

z + 17

Mila [183]3 years ago
3 0
Z+17 if I’m not mistaking
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What is the quotient in simplified form ? Need help please
Gre4nikov [31]

Answer:

  -\dfrac{4a^{12}b^8}{5}

Step-by-step explanation:

The applicable rules of exponents are ...

  \dfrac{1}{a^{-b}}=a^b\\\\a^ba^c=a^{b+c}

Your ratio simplifies to ...

  \dfrac{-8a^8b^{-2}}{10a^{-4}b^{-10}}=-\dfrac{4}{5}a^{8-(-4)}b^{-2-(-10)}\\\\=-\dfrac{4a^{12}b^8}{5}

_____

Please note that simplifying the constants, -8/10 to -4/5, eliminates half the answer choices. Since exponents are only multiplied when a power is raised to a power, you can pretty much eliminate the second choice as being unreasonable. ((a^b)^c = a^(bc))

8 0
4 years ago
Translate the following: 9 less than three times a number 2. 6 times the quotient of x and two 3. 4 times the difference of n an
KATRIN_1 [288]
(9-3)_
6 (x÷2)
4 (n-8)

I think this is right
6 0
3 years ago
Find the product of the<br>L.C.M and H.C.F of 8 and 12.​
V125BC [204]

Answer:

<h3>96 is the answer. </h3>

Step-by-step explanation:

<h3>The L.C.M of 8 and 12 is 24</h3><h3>The H.C.F of 8 and 12 is 4</h3>

<h3>We multiply </h3><h3>24 *4 we shall get 96 as the answer. </h3><h3 />
3 0
3 years ago
Read 2 more answers
The frequency n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m.
Nuetrik [128]

Answer:

n ∝ (1/l) (√F/m)

Step-by-step explanation:

Since frequency, n of vibration of a stretched string is a function of its tension F, the length l and the mass per unit length m, and has dimensions [T]⁻¹ where T = time, its dimension must be equal to that of the combination of F, L and m

Since n = f(F,L,m)

n = kFᵃLᵇmˣ

dimension of n = (dimension of F)ᵃ × (dimension of L)ᵇ × (dimension of m)ˣ

Since n = frequency, dimension of n = [T]⁻¹

F = Force, dimension of F = [M][L][T]⁻²

Also, L = length, dimension of L = [L] and

m = mass per unit length, dimension of m = [M][L]⁻¹

So, n = FᵃLᵇmˣ

[T]⁻¹  = ([M][L][T]⁻²)ᵃ( [L] )ᵇ([M][L]⁻¹)ˣ

[T]⁻¹  = ([M]ᵃ[L]ᵃ[T]⁻²ᵃ[L]ᵇ([M]ˣ[L]⁻ˣ

[M]⁰[L]⁰[T]⁻¹  = ([M]ᵃ ⁺ˣ)([L]ᵃ ⁺ ᵇ ⁻ˣ)([T]⁻²ᵃ)

equating the exponents on both sides, we have

a + x = 0  (1 )⇒ x = -a

a + b - x = 0 (2) ⇒

-2a = -1 (3) ⇒ a = 1/2

Substituting x into 2, we have

a + b -(-a) = 0

a + b + a = 0

2a + b = 0

b = -2a

b = -2 (1/2)

b = -1

x = -a = -1/2

So, substituting the variables into n, we have

n = kFᵃLᵇmˣ

n = kFᵃLᵇmˣ

n = kF^{\frac{1}{2} }L^{-1}m^{-\frac{1}{2} }

n = k/l(√F/m)

n ∝ (1/l) (√F/m)

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3 years ago
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