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3 years ago

How do you find the diameter of a circle please explain

Mathematics

1 answer:

Doss [256]3 years ago

3 0

You find the radius and you multiply it by 2, that is diameter because raidus is half of diameter

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Please help I’m having trouble with this question!

mezya [45]

ok thanks for letting me know you need help

4 0

2 years ago

4. Two friends, Sequoia and Raven, sold organic chap sticks at the local market.

Nina [5.8K]

Answer:

Sequoia sold 2 chap sticks while Raven sold 4.

Step-by-step explanation:

Since two friends, Sequoia and Raven, sold organic chap sticks at the local market, and Sequoia sold her chap sticks for $ 4 and Raven sold hers for $ 3 each, and in the first hour, their total combined sales were $ 20, if in the first hour, the friends sold a total of 6 chap sticks between them, to determine the number of chap sticks each of the friends sold during this time the following calculation must be performed:

6 x 4 + 0 x 3 = 24

5 x 4 + 1 x 3 = 23

2 x 4 + 4 x 3 = 20

Therefore, Sequoia sold 2 chap sticks while Raven sold 4.

7 0

2 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

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

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