All categories

4 years ago

Bottled apple juice is sold in four different different sizes. Which is the best to buy?

Mathematics

1 answer:

lbvjy [14]4 years ago

4 0

Answer:

the bigger one ;)

Step-by-step explanation:

You might be interested in

How do I find x using a mathematical operation?

Svetradugi [14.3K]

I got this. Bottom of the triangle. that flat line is 180 degrees. you have an angle of 120, leaving an angle of 60. if you look at the top, that angle matches the other. so you have 65 degrees on one side, and 60 on the other. so 180-65-60=55. so X=55

6 0

3 years ago

Can someone help me with this

Vilka [71]

Put 0, 18, 0 x

- Harvard Math

5 0

3 years ago

If PA = 6, PD = 4 and BE = 5. Find PE A) 4 B) 5 C) 6

lana [24]

Answer:

Option A) 4

Step-by-step explanation:

we know that

Applying the Intersecting Secant-Tangent Theorem

$PA^{2}=(BE+PE)( PE)$

substitute the given values

$6^{2}=(5+PE)( PE)\\ \\36=PE^{2}+5PE\\ \\PE^{2}+5PE-36=0$

Solve the quadratic equation by graphing

The solution is PE=4 units

see the attached figure

4 0

3 years ago

Divide 9x ^ 3 + 18x ^ 2 - 13x + 5 by 3x - 1 using division and write the division in the form P = DQ + R

lana66690 [7]

Answer:

$\frac{9x^3\:+\:18x\:^2\:-\:13x\:+\:5}{3x-1}=9x^3+18x^2-13x+5$

Step-by-step explanation:

DIVISION ALGORITHM: If $p(x)$ and $d(x)\neq 0$ are polynomials, and the degree of $d(x)$ is less than or equal to the degree of $f(x)$ , then there exist unique polynomials $q(x)$ and $r(x)$ , so that

$\frac{p(x)}{d(x)} =q(x)+\frac{r(x)}{d(x)}$

and so that the degree of $r(x)$ is less than the degree of $d(x)$ .

To find $\frac{9 x^{3} + 18 x^{2} - 13 x + 5}{3 x - 1}$ you must:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}9x^3+18x^2-13x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{9x^3}{3x}=3x^2$

$\mathrm{Quotient}=3x^2$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}3x^2:\:9x^3-3x^2\\\mathrm{Subtract\:}9x^3-3x^2\mathrm{\:from\:}9x^3+18x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\$

$\mathrm{Remainder}=21x^2-13x+5$

Therefore,

$\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+\frac{21x^2-13x+5}{3x-1}$

$\mathrm{Divide}\:\frac{21x^2-13x+5}{3x-1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}21x^2-13x+54\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{21x^2}{3x}=7x\\\\\mathrm{Quotient}=7x$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}7x:\:21x^2-7x\\\mathrm{Subtract\:}21x^2-7x\mathrm{\:from\:}21x^2-13x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=-6x+5$

Therefore,

$\frac{21x^2-13x+5}{3x-1}=7x+\frac{-6x+5}{3x-1}\\\\\frac{9x^3+18x^2-13x+5}{3x-1}=3x^2+7x+\frac{-6x+5}{3x-1}$

$\mathrm{Divide}\:\frac{-6x+5}{3x-1}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x+5\\\mathrm{and\:the\:divisor\:}3x-1\mathrm{\::\:}\frac{-6x}{3x}=-2\\\\\mathrm{Quotient}=-2$

$\mathrm{Multiply\:}3x-1\mathrm{\:by\:}-2:\:-6x+2\\\mathrm{Subtract\:}-6x+2\mathrm{\:from\:}-6x+5\mathrm{\:to\:get\:new\:remainder}\\\\\mathrm{Remainder}=3$