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nalin [4]
2 years ago
12

Math homework help exponents

Mathematics
1 answer:
Len [333]2 years ago
6 0

Answer + Step-by-step explanation:

\frac{5^{10}}{5^{5}} =5^{10-5} = 5^5

\text{used property :} \  \  \  \frac{a^{m}}{a^{n}} =a^{m-n}

_______________________

(4^8)^3 = a^{8 \times 3}=a^{24}

\text{used property :} \  \  \  (a^n)^m = a^{n \times m}

_______________________

10^{-4}=\frac{1}{10^{4}} \  \text{not} \  \frac{1}{4^{10}}

_______________________

15⁶ × 15³ = 15⁶⁺³ = 15⁹

_______________________

<u>Recall</u> : If a ≠0  ⇒ a⁰ = 1

Then

Since 6⁸ ≠ 0  ⇒ (6⁸)⁰ = 1  

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How do I solve bh+hr=25 when solving for b
Alexxx [7]
Bh + hr = 25

you can factor out an h from the left side of your equation:

h(b + r) = 25 ... divide both sides by h
b + r = (25/h) ... subtract r
b = (25/h) - r
3 0
3 years ago
Read 2 more answers
Which Iink between two elements could you remove from the relation so that becomes a function?
OLEGan [10]

Answer:

Either (3,b) or (3,e) must be removed to form the above relation to be function.

Step-by-step explanation:

Given the relation between the elements of domain and range. we have to remove the link between two elements so that relation becomes a function.

A function is a relation in which each element in domain set maps into only one element of range set i.e one element can never have two images in range set.

Here, the element 3 in domain set maps into two element b and e in range set. Therefore, to form a function one link must be removed either

f(3)=b or f(3)=e

Hence, either (3,b) or (3,e) must be removed to form the above relation to be function.

Read more on Brainly.com - brainly.com/question/5314192#readmore

4 0
3 years ago
Given the linear equations below determine if the lines are parallel, perpendicular, or intersecting.
IgorC [24]

Answer: They would be intersecting lines

Step-by-step explanation:

6 0
4 years ago
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Which equation represents the line that passes through (-6, 7) and (-3, 6)?
9966 [12]

Answer:

y = -3x - 11

Step-by-step explanation:

(y - y1)/(x - x1) = (x2 - x1)/(y2 - y1)

(y - 7)/(x + 6) = (-3 + 6)/(6 - 7)

(y - 7)/(x + 6) = 3/-1 = -3

y - 7 = -3(x + 6)

y - 7 = -3x - 18

y = -3x -18 - 7

y = -3x - 11

4 0
4 years ago
Find all rational zeros of the polynomial
rjkz [21]

x^3+x^2-5x+3=x^3+3x^2-2x^2-6x+x+3\\\\=(x^3-2x^2+x)+(3x^2-6x+3)=x(x^2-2x+1)+3(x^2-2x+1)\\\\=(x^2-2x+1)(x+3)=\underbrace{(x^2-2(x)(1)+1^2)}_{(a-b)^2=a^2-2ab+b^2}(x+3)=(x-1)^2(x+3)\\\\P(x)=0\iff(x-1)^2(x+3)=0\iff(x-1)^2=0\ \vee\ (x+3)=0\\\\x-1=0\ \vee\ x-1=0 \vee\ x+3=0\\\\x=1\ \vee\ x=1\ \vee\ x=-3\\\\Answer:\ \boxed{x_1=1,\ x_2=1,\ x_3=3}

4 0
3 years ago
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