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2 years ago

HELP ASAP ILL GIVE A BRAINLISET

Mathematics

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galina1969 [7]2 years ago

8 0

Answer:

am an arts student my friend

Step-by-step explanation:

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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

Each year, more than 2 million people in the United States become infected with bacteria that are resistant to antibiotics. In p

Bingel [31]

Answer:

Null hypothesis:H₀:- There is no significant difference between in drug resistance between the two states

Alternative Hypothesis :H₁:

There is significant difference between in drug resistance between the two states

B)

The calculated value Z = 2.7261 > 2.054 at 0.02 level of significance

Rejected H₀

There is a significant difference in drug resistance between the two states.

P - value = 0.0066

P - value = 0.0066 < 0.02

D)

1) Reject H₀

There is a significant difference in drug resistance between the two states.

Step-by-step explanation:

Step(i):-

Given first sample size n₁ = 174

Suppose that, of 174 cases tested in a certain state, 11 were found to be drug-resistant.

First sample proportion

$p_{1} = \frac{x_{1} }{n_{1} } = \frac{11}{174} = 0.0632$

Given second sample size n₂ = 375

Given data Suppose also that, of 375 cases tested in another state, 7 were found to be drug-resistant

Second sample proportion

$p_{2} = \frac{x_{2} }{n_{2} } = \frac{7}{375} = 0.0186$

Step(ii):-

Null hypothesis:H₀:- There is no significant difference between in drug resistance between the two states

Alternative Hypothesis :H₁:

There is significant difference between in drug resistance between the two states

Test statistic

$Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

Where

$P = \frac{n_{1}p_{1} +n_{2} p_{2} }{n_{1} +n_{2} }$

$P = \frac{174 (0.0632) + 375 (0.0186) }{174+375 } = \frac{17.9718}{549} = 0.0327$

Q = 1 - P = 1 - 0.0327 = 0.9673

Step(iii):-

Test statistic

$Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$

$Z = \frac{0.0632-0.0186 }{\sqrt{0.0327 X 0.9673(\frac{1}{174 }+\frac{1}{375 } ) } }$

Z = 2.7261

Level of significance = 0.02 or 0.98

The z-value = 2.054

The calculated value Z = 2.7261 > 2.054 at 0.02 level of significance

Reject H₀

There is a significant difference in drug resistance between the two states.

P- value

P( Z > 2.7261) = 1 - P( Z < 2.726)

= 1 - ( 0.5 + A (2.72))

= 0.5 - 0.4967

= 0.0033

we will use two tailed test

2 P( Z > 2.7261) = 2 × 0.0033

= 0.0066

P - value = 0.0066 < 0.02

Reject H₀

There is a significant difference in drug resistance between the two states.

8 0

2 years ago

The daltons completed 30% of their 2140-mile trip the first day

Ugo [173]

To figure out how many miles the Daltons traveled the first day you times 2140 miles and 30 percent.
So 2140 x .30 = 642
They traveled 642 miles the first day.

To figure out how many miles they still have to travel you subtract your original miles (2140) from how many miles they have already traveled (642).
So 2140 - 642 = 1498
They have 1498 miles to still travel.

4 0

2 years ago

Read 2 more answers

What is the value of m? m3 = 64

uranmaximum [27]

Answer

m=4

Step-by-step explanation:

3 0

3 years ago

lydia created a rectangle shape design that is 12 inches wide and 15 inches long she plans to make an enlargement of the design

Romashka [77]

Answer: 1200 inches long.

Step-by-step explanation:

the enlargement went from 1 inches wide to 80 inches wide.So which means it was being multiply by 80.

so you have to multiply 15 by 80 to find length of the enlargement or how long the enlargement of the design will be.

15 * 80=1200

6 0

3 years ago