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2 years ago

Which graph shows a linear fution

Mathematics

1 answer:

Goryan [66]2 years ago

7 0

I need to see the graphs, can you attach a image of them..

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The length is 2 feet longer than the width. The perimeter is 36 feet, what is the length?

SOVA2 [1]

Answer:

The length would be 10ft and the width would be 8ft

Step-by-step explanation:

For the purpose of this, we'll set the width as x. We can then define the length as x + 2 since we know it is 2 ft longer than the width. Now we can use those along with the perimeter formula to solve for the width.

P = 2l + 2w

36 = 2(x + 2) + 2(x)

36 = 2x + 4 + 2x

36 = 4x + 4

32 = 4x

8 = x

Now since we know that the width is 8ft, we can add 2ft to it to get the length, which would be 10ft.

4 0

3 years ago

Estimate the quotient: 1628 ÷ 8

Pachacha [2.7K]

Answer:

203.5

Step-by-step explanation:

Please mark brainliest.

4 0

3 years ago

Read 2 more answers

What can you tell about a triangle when given three side lengths?

Genrish500 [490]

Check whether it is possible to have a triangle with the given side lengths for example 7,9,13 then add any two sides and see if it is greater than the other side. Example the sum of 7 and 9 is 16 and 16 is greater than 13

3 0

3 years ago

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

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

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

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

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

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

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

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

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

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,

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

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

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

So,

Equation [A] becomes

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

So, the polynomial equation becomes

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

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