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4 years ago

9

Help me find the area of this hexagon.

1 answer:

Alika [10]4 years ago

6 0

Answer:

70

Step-by-step explanation:

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A 12 pack of diet soda has dimensions 4.75in and 28.5in whats the volume of the 12 pack round to the nearest ten

ExtremeBDS [4]

Answer:

$Volume = 643.0\ in^3$

Step-by-step explanation:

Given

Dimension: 4.75in, 4.75in and 28.5in

<em>[The complete dimension is missing from the question]</em>

Required

Calculate the volume

The volume is calculated as thus:

$Volume = 4.75in * 4.75in * 28.5in$

$Volume = 643.03125in^3$

$Volume = 643.0\ in^3$ --- approximated

6 0

3 years ago

Alan is 14 years old.This is twice as old as his brother James is.How old is James?

Vladimir [108]

His brother would be 7 years old.

6 0

4 years ago

Read 2 more answers

Who can help me with this algebra thanks for your help guys.

hjlf

31. +
34. -
To figure out more, just solve the known problem, and then go over what the answer would be to the unknown one until you find the answer that makes it true.

6 0

3 years ago

A fifth-degree polynomial equation with rational coefficients has the roots 3, 8i, and 7- radical 5. Which are also roots of the

Sliva [168]

Using <u>complex numbers and radicals</u>, it is found that the correct option is C.

When a complex number is a root of a polynomial function, it's conjugate also has to be.

Hence, since 8i is a root of the function, -8i is also a root.

The same is valid for radicals, as they are in the $\pm$ format.

Since $7 - \sqrt{5}$ is a root, $7 + \sqrt{5}$ is also a root.

Hence, option C is correct.

For more on <u>complex numbers and radicals</u>, you can check brainly.com/question/25173944

8 0

2 years ago

Foil: (x+9)(3x-4)

QveST [7]

QUESTION 1

The given binomial is;

$(x+9)(3x-4)$

First terms are multiplied: $x\times 3x=3x^2$

Outside terms are multiplied: $x\times -4=-4x$

Inside terms are multiplied: $9\times 3x=27x$

Last terms are Multiplied: $9\times -4=-36$

This gives us;

$=3x^2-4x+27x-36$

$=3x^2+23x-36$

QUESTION 2

We want to factor

$6r^5s+4r^4s^2-8r^2s$

The HCF is $2r^2s$

We factor to get;

$2r^2s(3r^3+2r^2s-4)$

QUESTION 3;

$x^2+x-2=0$

Split the middle term;

$x^2+2x-x-2=0$

Factor

$x(x+2)-1(x+2)=0$

$(x+2)(x-1)=0$

$(x+2)=0,(x-1)=0$

$x=-2,x=1$

The solutions are;

$x=-2$ and $x=1$

These are the x-intercepts of the graph of the function

$f(x)=x^2+x-2$

3 0

4 years ago