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3 years ago

5

What is the value of the 4 in 542

1 answer:

Juli2301 [7.4K]3 years ago

8 0

It is in the tens place so it would be 40

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Could someone check this or correct me please?

Greeley [361]

I agree. !!!!!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago

Read 2 more answers

Help appreciated on question in image!<br> Thanks:)

Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

As

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

As

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$

$\mathrm{Solve\:}\:x+7=0:\quad x=-7$

$\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i$

$\mathrm{The\:solutions\:are}$

$x=-1,\:x=-7,\:x=i,\:x=-i$

Keywords: polynomial equation

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5 0

3 years ago

Read 2 more answers

Based on the picture..Find the area in terms of xFind the perimeter in terms of x

Vlad1618 [11]

The area of a rectangle is given by the product of its dimensions.

If this rectangle has a length of 2x + 7 and a width of x + 4, its area is:

$\begin{gathered} \text{Area}=\text{length}\cdot\text{width} \\ \text{Area}=(2x+7)(x+4)_{} \\ \text{Area}=2x^2+8x+7x+28 \\ \text{Area}=2x^2+15x+28 \end{gathered}$

The perimeter is the sum of all side lengths, so we have:

$\begin{gathered} \text{Perimeter}=(2x+7)+(x+4)+(2x+7)+(x+4) \\ \text{Perimeter}=6x+22 \end{gathered}$

6 0

1 year ago

What is the answer and why?

Sati [7]

A is the answer, because 3/5=.60 and 37.5/62.5=60

6 0

3 years ago

Find the length of the line segment whose endpoints are (-5,6) and (6,6)

umka21 [38]

Answer: 11 units

Step-by-step explanation:

Use the distance formula: $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

$(x_{1}, y_{1})=(-5, 6)\\(x_{2}, y_{2})=(6, 6)\\\\\sqrt{(6-(-5)^{2}+(6-6)^{2}} =\sqrt{(11)^{2}+(0)^{2}} =\sqrt{121+0}=\sqrt{121}=11$

5 0

3 years ago