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3 years ago

The sum of two numbers is 28 and there quotient is 3 what are the numbers

Mathematics

1 answer:

Zepler [3.9K]3 years ago

5 0

54 and 21 and thats how i got it lol

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WILL GIVE BRAINLIEST ASAPP!!!!!

Mademuasel [1]

Step-by-step explanation:

Applying rules of exponents to solve the given problems;

4^3 x 4^5 =

5^8 ÷ 5^-2 =

(6^3 ) ^ 4 =

For these problems, the applicable rules of exponents are;

aᵇ x aⁿ = aᵇ⁺ⁿ

aᵇ ÷ aⁿ = aᵇ⁻ⁿ

(aᵇ)ˣ = aᵇˣ

For the first problem; 4³ x 4⁵

aᵇ x aⁿ = aᵇ⁺ⁿ

4³ x 4⁵ = 4³⁺⁵ = 4⁸

Second problem: aᵇ ÷ aⁿ = aᵇ⁻ⁿ

5⁸ ÷ 5⁻² = 5⁸⁻⁽⁻²⁾ = 5⁸⁺² = 5¹⁰

Third problem; (aᵇ)ˣ = aᵇˣ

(6³)⁴ = 6³ˣ⁴ = 6¹²

5 0

3 years ago

Read 2 more answers

The formula for the sum of an infinite geometric series, S=a1/1-r, may be used to convert

schepotkina [342]

Answer:

D. a1=23/100, r=1/100

Step-by-step explanation:

The repeating fraction can be written as the sum ...

$0.\overline{23}=0.23+0.0023+0.000023+\dots$

The first term is a1 = 0.23 = 23/100, and each successive term is shifted 2 decimal places to the right, so is multiplied by the common ratio r=1/100.

7 0

3 years ago

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....................

Murljashka [212]

Answer:

Equal to each other

Step-by-step explanation:

I hope that helped!

4 0

3 years ago

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Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

If a and b are positive numbers, find the maximum value of f(x) = xa(7 â x)b on the interval 0 â¤ x â¤ 7.

algol [13]

Given $f(x)=x^a(7-x)^b$ on the interval 0 ≤ x ≤ 7, for maximum value, f'(x) = 0.

$f'(x)=0 \\ \\ \Rightarrow -bx^{a}(7-x)^{b-1}+ax^{a-1}(7-x)^b=0 \\ \\ \Rightarrow ax^{a-1}(7-x)^b=bx^{a}(7-x)^{b-1} \\ \\ \Rightarrow a(7-x)=bx \\ \\ \Rightarrow (a+b)x=7a \\ \\ \Rightarrow x= \frac{7a}{a+b}$

8 0

3 years ago