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Anni [7]
2 years ago
5

Rewrite each multiplication or division expression using a base and an exponent

Mathematics
1 answer:
brilliants [131]2 years ago
6 0

Using the laws of indices , 4⁵ ÷ 4² can rewritten as 4³.

<h3>What is the result of the division in exponent form?</h3>

Given the expression; 4⁵ ÷ 4²

To perform the division operation, we use one of the laws of indices;

n^a/n^b = n^{a-b}

Given that;

  • n = 4
  • a = 5
  • b = 2

Now, we apply the laws of indices ;

4⁵ ÷ 4² = 4⁵⁻² = 4³

Therefore, using the laws of indices, 4⁵ ÷ 4² can rewritten as 4³.

Learn more about exponents here: brainly.com/question/15993626?

#SPJ1

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Define $g$ by $g(x)=5x-4$. If $g(x)=f^{-1}(x)-3$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $5a+5b$.
Slav-nsk [51]

Answer:

<h2>2</h2>

Step-by-step explanation:

Given g(x)=5x-4 and g(x)=f^{-1}(x)-3

Substituting g(x) =  5x-4 into the second equation we have;

5x-4 =  f^{-1}(x)-3

f^{-1}(x) = 5x-4+3

f^{-1}(x) = 5x-1

To get f(x), let us first make y to be equal  f^{-1}(x)

y = 5x-1

expressing x in terms of y to get f(x), we have;

5x = y+1

x = y/5+1/5

replacing y with x, we will have;

y = x/5 + 1/5

F(x) = x/5 + 1/5

Comparing x/5 + 1/5 with ax+b, a = 1/5 and b = 1/5

5a + 5b = 5(1/5)+ 5(1/5)

5a+5b = 1+1

5a+5b = 2

8 0
3 years ago
4.1&lt;7 as a fraction. pls workout and ill mark you brainliest
labwork [276]

Answer:

4/7

Step-by-step explanation:

hope this helps

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3 years ago
Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
julia-pushkina [17]

Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0
3 years ago
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