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Anastasy [175]
3 years ago
8

A tube of toothpaste is marked $3.92 for 8 ounces. What is the price per ounce?

Mathematics
2 answers:
Alisiya [41]3 years ago
5 0
Find one ounce $3.92/8=$0.49
$0.49 per ounce
myrzilka [38]3 years ago
4 0

Answer:

$0.49 hope this helps you out

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Hey y’all I don’t understand, can you help me pls ASAP and show steps if u could
Shkiper50 [21]

Answer:

(x, y) = (-2, -6)

Step-by-step explanation:

Learn more at brainly.com/question/15168004

5 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
Evaluate x = 4 and y=5.<br>X? + 2(x + y)​
777dan777 [17]

4+2(4+5)

4+2*4+2*5=

4+8+10=

22

6 0
3 years ago
Given that f(x)=x-9,g(x) = 3x^2- 2x + 5, and h(x)=-6x, find each function.
Zina [86]

hi

f(x) = x-9  ;  

g(x) = 3x²-2x +5  

h(x) = -6x  

A :  f+g =  x-9  +3x²-2x+5  

     f+g =  3x²-x-4  

b :  h-g  =  -6x - 3x²+2x-5

     h-g =    -3x²-4x-5

8 0
3 years ago
Find the derivative with respect to x of y = (3x + x^2)^5
RSB [31]
<h2>The required "option A) 5(3x + x^2)^{4} (3+2x)" is correct.</h2>

Step-by-step explanation:

We have,

y = (3x + x^2)^5          ..... (1)

To find, \dfrac{dy}{dx} = ?

Differentiating equation (1) w.r.t. 'x', we get

\dfrac{dy}{dx}= \dfrac{d[(3x + x^2)^5]}{dx}

⇒ \dfrac{dy}{dx}=5(3x + x^2)^{5-1} \dfrac{d(3x + x^2)}{dx}

[ ∵ y=x^{n} ⇒ \dfrac{dy}{dx}=nx^{n-1}]

⇒ \dfrac{dy}{dx}=5(3x + x^2)^{4} (3(1) + 2x^{2-1})

⇒ \dfrac{dy}{dx}=5(3x + x^2)^{4} (3+2x)

Thus, the required "option A) 5(3x + x^2)^{4} (3+2x)" is correct.

3 0
3 years ago
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