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katovenus [111]
3 years ago
8

Please help . I’ll mark you as brainliest if correct!!!!!

Mathematics
1 answer:
topjm [15]3 years ago
6 0

Answer:

Step-by-step explanation:

hi

pls find below

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What is the y-intercept of the function y = –2(2)^x+ 2?
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so is 2
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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

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3 years ago
Kate bought $23.40 worth of two types of bird seed. Thistle bird seed sells for $1.60 per pound and wild bird seed sells for $0.
earnstyle [38]

Answer:

Ax+By=C and 7.5 lb of Thistle

Step-by-step explanation:

6 0
3 years ago
Holly says 4.131131113... is a rational number. Which of the following best describes whether Holly is correct and why?? a.Holly
Leni [432]

Answer:

A. Holly is correct. Repeating decimals are always rational numbers.

Step-by-step explanation:

Rational number is the subset of:

  • Whole number (Positive integer + 0)
  • Natural number
  • Negative integers
  • Fractions
  • Decimals (Whether positive or not, whether terminating or non terminating)

Hope this helps ;) ❤❤❤

7 0
3 years ago
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If $-2 < x \le 3,$ then find all possible values of $5x + 1.$ Give your answer in interval notation PLEASE ANSWER THUS PLEASE
exis [7]

Answer:

(-9,16]

Step-by-step explanation:

we know that

The domain of x is equal to the interval--------> (-2,3]

-2< x \leq 3

All real numbers greater than -2 and less than or equal to 3

Let

y=5x+1

For x=3

y=5x+1\\y=5(3)+1=16

For x>-2

y >5(-2)+1\\y > -9

so

The range of the function is the interval--------> (-9,16]

5 0
3 years ago
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