All categories

3 years ago

Which of the following statements is true?

Mathematics

2 answers:

N76 [4]3 years ago

6 0

Every square is a rhombus

sashaice [31]3 years ago

5 0

Answer:

every square is a rhombus

Step-by-step explanation:

You might be interested in

Please help me!!!!!!

UkoKoshka [18]

Answer:

Step-by-step explanation:

because u have 5 potatoe bags and it .............

6 0

3 years ago

Please help quick -2=2+ v/4

cluponka [151]

The answer to this problem is v=-16

5 0

3 years ago

Two boys and three girls are auditioning to play the piano for a school production. Two students will be chosen, one as the pian

Sergio039 [100]

The answer is 0.24

To calculate this, a multiplication rule is used. The multiplication rule calculates the probability that both of two events will occur. In this method, the probabilities of each event are multiplied. Here we have two events:

1. The probability that the pianist will be a boy,

2. The probability that the alternate will be a girl.

So, let's calculate these probabilities:

There are in total 5 children:

2 boys + 3 girls = 5 children.

1. The probability that the pianist will be a boy is 2 boys out of 5 children, which is 2/5.

2. The probability that the alternate will be a girl is 3 girls out of 5 children, which is 3/5.

The probability that the pianist will be a boy and the alternate will be a girl is:

2/5 × 3/5 = 6/25

6/25 × 4/4 = 24/100 = 0.24

3 0

3 years ago

Read 2 more answers

What is the answer to 37+45?

poizon [28]

The answer is 82 because 5+7=12 then 3+4=7 plus the 1 = 82

6 0

3 years ago

Read 2 more answers

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