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faltersainse [42]

3 years ago

10

A nearby pond has 5 frogs, and the population doubles every year. The inequality 5(2)^t>75, where t is the number of years, m

odels when the population of frogs will be greater than 75.
Based on the inequality, when will the population of frogs in the pond be greater than 75?

t>4 years
t>7.5 years
t>2 years
t>1 year

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer: t > 4 years

Explanation: This was the only logical answer, and I’ve taken the quiz.

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What is.. 4x-2=(x+3)+2

Contact [7]

Answer:

x = 7/3

Step-by-step explanation:

4x - 2 = x + 3 + 2

4x - 2 = x + 5

Rearranging the terms, we get,

4x - x = 5 + 2

3x = 7

x = 7/3

Hence, x = 7/3.

5 0

3 years ago

Find all solutions to the following quadratic equations, and write each equation in factored form.

dexar [7]

Answer:

(a) The solutions are: $x=5i,\:x=-5i$

(b) The solutions are: $x=3i,\:x=-3i$

(c) The solutions are: $x=i-2,\:x=-i-2$

(d) The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) The solutions are: $x=1$

(g) The solutions are: $x=0,\:x=1,\:x=-2$

(h) The solutions are: $x=2,\:x=2i,\:x=-2i$

Step-by-step explanation:

To find the solutions of these quadratic equations you must:

(a) For $x^2+25=0$

$\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25$

$\mathrm{Simplify}\\x^2=-25$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-25},\:x=-\sqrt{-25}$

$\mathrm{Simplify}\:\sqrt{-25}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\\\\\sqrt{-25}=\sqrt{-1}\sqrt{25}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\\sqrt{-25}=\sqrt{25}i\\\\\sqrt{-25}=5i$

$-\sqrt{-25}=-5i$

The solutions are: $x=5i,\:x=-5i$

(b) For $-x^2-16=-7$

$-x^2-16+16=-7+16\\-x^2=9\\\frac{-x^2}{-1}=\frac{9}{-1}\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{-9},\:x=-\sqrt{-9}$

The solutions are: $x=3i,\:x=-3i$

(c) For $\left(x+2\right)^2+1=0$

$\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x+2=\sqrt{-1}\\x+2=i\\x=i-2\\\\x+2=-\sqrt{-1}\\x+2=-i\\x=-i-2$

The solutions are: $x=i-2,\:x=-i-2$

(d) For $\left(x+2\right)^2=x$

$\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4$

$x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}$

$x_1=\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}+i\frac{\sqrt{7}}{2}\\\\x_2=\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}-i\frac{\sqrt{7}}{2}$

The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) For $\left(x^2+1\right)^2+2\left(x^2+1\right)-8=0$

$\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5$

$\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0$

$u_1=\frac{-4+\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad 1\\\\u_2=\frac{-4-\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad -5$

$\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=\sqrt{5}i,\:x=-\sqrt{5}i$

The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) For $\left(2x-1\right)^2=\left(x+1\right)^2-3$

$\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0$

$\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:3}\\x=\frac{-\left(-6\right)}{2\cdot \:3}\\x=1$

The solutions are: $x=1$

(g) For $x^3+x^2-2x=0$

$x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2$

The solutions are: $x=0,\:x=1,\:x=-2$

(h) For $x^3-2x^2+4x-8=0$

$x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i$

The solutions are: $x=2,\:x=2i,\:x=-2i$

3 0

3 years ago

plz i need help with this its due tmrw!! (also my other most recent question is #6 of this and it would be nice for answers for

kipiarov [429]

7. <u>You have $367.50 after two years.</u>

<em>Start by converting 2.5% into a decimal (divide by 100) and multiplying by 350 to find the rate of interest per year.</em>

<em>350(0.025) = 8.75</em>

<em>Since it's for two years, multiply by two. </em>

<em>8.75 x 2 = 17.5.</em>

<em>Add it on to he original, and we have</em>

<em>350 + 17.5 = 367. 5, or $367.50 when converted back to money. </em>

8. <u>The annual interest rate is 2.4%</u>

<em>find the interest rate per year: </em>

<em>120/2.5 = 48 dollars per year. this is the interest amount, we want to find the interest rate. To do this, find what % of 2000 that 48 is equal to. </em>

<em>Set up a system of equations and cross multiply.</em>

<em /> $\frac{48}{2000} = \frac{x}{100}$ <em />

<em>2000x = 48(100) > 2000x = 4800</em>

<em>2000</em><em>/2000</em><em>x = 4800</em><em>/2000 > </em><em>x = 2.4</em>

<em>So, the interest rate is 2.4%. </em>

9. <u>the interest paid is $300 after six months, and $600 after a year.</u>

<em>Find the interest rate, similar to problem 7. </em>

<em>3000 x 0.2 = 600. This means the interest paid is $600 a year. In six months, the total will be half, or $300.</em>

10. <u>four years.</u>

<em>Find interest rate. </em>

<em>200(0.035) = $7/year</em>

<em>Remember that value. Subtract needed from current.</em>

<em>228 - 200 = $28. </em>

<em>So, we have an interest rate of $7 a year and we need $28. Normally, we'd solve using an expression, but in this case we can use simple multiplication. Knowing that 7 x 4 = 28, We can decide that it will take four years. </em>

5 0

3 years ago

Beth is selling tickets for a dinner at her school.

iren [92.7K]

Answer:

D

Step-by-step explanation:

check each child answer and multiply by 4.50

check each adult answer and multiply by 6.00

add both answers up

and see which one it is

its not A, B, or C

5 0

3 years ago

Read 2 more answers

Please someone help with my math

Pepsi [2]

Answer:

1. .31

2. .68

3. .08

4. true???

5. 2.83

6. .75

7. true again??

8. .09

9. 4.55

10 .10

11. 9.33

12. .92

13. 14.16

Step-by-step explanation:

i didnt really understand 4 and 7 sooo true?

6 0

4 years ago