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3 years ago

14

Daddy needs to make 16 costumes for the school play.Each costume requires 2 1/4 yards of material. How many yards of material wi

ll he need

1 answer:

sveticcg [70]3 years ago

5 0

7 costumes

I found this by dividing the two numbers and when divided the answer is 7 and 1/9 but you cant have 1/9th of a costume so it would only be 7

Hope this helps

If any questions leave a comment

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Kitty [74]

Your answer is b. hope this helps :)

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3 years ago

Select an equation that passes through the point (5,7) and is perpendicular to the line x=4

alexandr1967 [171]

y = 7

x = 4 is a vertical line parallel to the y- axis passing through all points with an x- coordinate of 4

A line perpendicular to x = 4 will be a horizontal line parallel to the x- axis with equation y = c where c is the value of the y- coordinate the line passes through.

line passes through (5 , 7) with y- coordinate 7

The equation is therefore y = 7

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The Pic. Shows it. Please show the steps.<br> Will MARK BRAINLIEST

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Step-by-step explanation:

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3 years ago

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1 ÷ 3 Express your answer as a fraction

Varvara68 [4.7K]

Answer:

1/3

Step-by-step explanation:

1 ÷ 3 = 1/3

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3 years ago

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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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2 years ago

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