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## How do you simplify 3/5×(-1/4 - 1/6) ÷ (-7/3 + 5/4)?

Meave60
Featured 2 months ago

3/5 xx (-1/4-1/6) -:(-7/3+5/4)=color(blue)(3/13

#### Explanation:

Simplify:

$\frac{3}{5} \times \left(- \frac{1}{4} - \frac{1}{6}\right) \div \left(- \frac{7}{3} + \frac{5}{4}\right)$

Follow the order of operations: parentheses, exponents, multiplication and division left to right, addition and subtraction left to right.

Simplify the parentheses first.

$\left(- \frac{1}{4} - \frac{1}{6}\right)$ must have a common denominator. The least common denominator (LCD) can be found by listing the multiples of $4$ and $6$.

$4 :$$4 , 8 , \textcolor{red}{12} , 16. . .$

$6 :$$6 , \textcolor{red}{12.} . .$

LCD$=$$\textcolor{red}{12}$

Multiply both fractions by a fraction equal to $1$, so that each will have the same denominator. For example, $\frac{5}{5} = 1$. This way, the values don't change.

$\left(- \frac{1}{4} \times \frac{\textcolor{t e a l}{3}}{\textcolor{t e a l}{3}} - \frac{1}{6} \times \frac{\textcolor{m a \ge n t a}{2}}{\textcolor{m a \ge n t a}{2}}\right)$

Simplify.

$\left(- \frac{3}{12} - \frac{2}{12}\right) = \left(- \frac{5}{12}\right)$

Return $\left(- \frac{5}{12}\right)$ to the original expression.

$\frac{3}{5} \times \left(- \frac{5}{12}\right) \div \left(- \frac{7}{3} + \frac{5}{4}\right)$

Simplify $\left(- \frac{7}{3} + \frac{5}{4}\right)$.

Follow the same procedure as with $\left(- \frac{1}{4} - \frac{1}{6}\right)$.

LCD$=$$12$

$\left(- \frac{7}{3} \times \frac{\textcolor{g r e e n}{4}}{\textcolor{g r e e n}{4}} + \frac{5}{4} \times \frac{\textcolor{red}{3}}{\textcolor{red}{3}}\right)$

Simplify.

$\left(- \frac{28}{12} + \frac{15}{12}\right) = - \frac{13}{12}$

Return $\left(- \frac{13}{12}\right)$ to the original equation.

$\frac{3}{5} \times \left(- \frac{5}{12}\right) \div \left(- \frac{13}{12}\right)$

Multiply $\frac{3}{5}$ and $- \frac{5}{12}$.

$- \frac{15}{60} \div \left(- \frac{13}{12}\right)$

Divide $- \frac{15}{60}$ and $- \frac{13}{12}$.

When dividing by a fraction, invert the fraction and multiply.

$- \frac{15}{60} \times - \frac{12}{13}$

Simplify.

$\frac{180}{780}$

Reduce by dividing the numerator and denominator by $60$.

$\frac{180 \div 60}{780 \div 60} = \frac{3}{13}$

## Four less than the product of a number and 8 is twice the sum of the number and 10. What is the number?

Jon
Featured 1 month ago

$x = 4$

#### Explanation:

Let the number be represented by some variable, let's call it $x$.
We can reword the question into an algebraic equation so we can solve for $x$ and find out what the number is.

"Four less than the product of the number and 8" is basically saying:

$8 x - 4$

We can let the word "is" in this question just be an equal sign, because we are saying this is that. In other words, this = that.

"Twice the sum of the number and 10" means:

$2 \left(x + 10\right)$

So, we now have a complete algebraic equation:

$8 x - 4 = 2 \left(x + 10\right)$

We can simplify this into simpler terms:

$8 x - 4 = 2 x + 20$

Since we are trying to find out what the $x$ is, it is best to get the terms with an $x$ on one side, and all other terms on the other side. To do this, subtract $2 x$ from both sides, and add $4$ to both sides to get:

$6 x = 24$

Now, we can find out what $x$ is by getting it all by itself.
Divide by 6 to do this:

$x = 4$

## How do you find the simplest radical form of 433?

George C.
Featured 1 week ago

Hmmm...

#### Explanation:

If you mean the simplest form of $\sqrt{433}$ then it is $\sqrt{433}$ since $433$ is a prime number.

If you mean the simplest expression involving a radical with value $433$, then you might choose $433 \sqrt{1}$ or possibly $\sqrt{187489}$ (since $433 \cdot 433 = 187489$).

Approximations

Since $433$ is a prime number, its square root cannot be simplified. In addition, it is not close to a square number or half way between two square numbers. So it is a little fiddly to get a rapidly convergent continued fraction expressing it.

Note that:

${20}^{2} = 400 < 433 < 441 = {21}^{2}$

So a reasonable approximation is somewhere between $20$ and $21$. Linearly interpolating, we find it is about $20 + \frac{33}{41} \approx 20.8 = \frac{104}{5}$

Then:

$433 - {\left(\frac{104}{5}\right)}^{2} = \frac{9}{25}$

Now:

$\sqrt{{a}^{2} + b} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$

So:

$\sqrt{433} = \frac{104}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \frac{\frac{9}{25}}{\frac{208}{5} + \ldots}}}$

$\textcolor{w h i t e}{\sqrt{104}} = \frac{1}{5} \left(104 + \frac{9}{208 + \frac{9}{208 + \frac{9}{208 + \ldots}}}\right)$

Hence, if we define a sequence recursively by:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 208 {a}_{n + 1} + 9 {a}_{n}\end{matrix}\right.$

then the ratio between successive terms rapidly converges to $5 \sqrt{433} + 104$

The first few terms are:

$0 , 1 , 208 , 43273 , 9002656$

So:

$5 \sqrt{433} \approx \frac{9002656}{43273} - 104 = \frac{4502264}{43273}$

So:

$\sqrt{433} \approx \frac{4502264}{5 \cdot 43273} = \frac{4502264}{216365} \approx 20.808652046$

## How do I solve the following?

Gió
Featured 4 weeks ago

I got:
Bluberry pie $4 Lemmon pie: $6

#### Explanation:

Call the costs:
blueberry pie $b$
lemmon pie $l$

we get:

$12 b + 2 l = 60$ for Perry
$6 b + 4 l = 48$ for Kristin

from the first equation we can isolate $l$ and write:
$l = - \frac{12}{2} b + \frac{60}{2}$
$l = - 6 b + 30$

substitute into the second for $l$ (in red):

$6 b + 4 \left(\textcolor{red}{- 6 b + 30}\right) = 48$
solve for $b$:
$6 b - 24 b + 120 = 48$
$18 b = 72$
$b = \frac{72}{18} = 4$

into the expression for $l$:

$l = - 6 \left(4\right) + 30 = - 24 + 30 = 6$

## Show that, if k is any real number, the equation x^2 + (k+1)x + k = 0 always has real roots. For what values of k are the roots equal?

Featured 1 week ago

See the explanation below

#### Explanation:

${x}^{2} + \left(k + 1\right) x + k = 0$ where $k \in \mathbb{R}$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(k + 1\right)}^{2} - 4 \cdot 1 \cdot k = {\left(k + 1\right)}^{2} - 4 k$

$= {k}^{2} + 2 k + 1 - 4 k$

$= {k}^{2} - 2 k + 1$

$= {\left(k - 1\right)}^{2}$

Therefore,

$\forall k \in \mathbb{R}$, ${\left(k - 1\right)}^{2} \ge 0$

$\iff$, $\Delta \ge 0$

Conclusion : As $\Delta \ge 0$, for all values of $k \in \mathbb{R}$, the roots of the quadratic equation are always real.

For the roots to be equal, $\Delta = 0$

Therefore,

${\left(k - 1\right)}^{2} = 0$

$\iff$, $k = 1$

When $k = 1$, the roots of the quadratic equation will be equal.

## How do you write a function rule for the area of a triangle with a base 10 cm less than 8 times the height. What is the area of the triangle when its height is 5 cm?

Nimo N.
Featured 1 week ago

$A \left(H\right) = 4 {H}^{2} - 5 H$
$A \left(5 c m\right) = 75 c {m}^{2}$

#### Explanation:

Area of a triangle with a base 10 cm less than 8 times the height.
Area of a triangle: $A = \frac{1}{2} \cdot B \cdot H$

Relationship between B and H:
$B = 8 \cdot H - 10$

It is unspecified whether the area should be in terms of its base or height, A(H) will be used, especially in light of the second part of the question.

So, substitute the expression for B into the area equation.
$A \left(H\right) = \frac{1}{2} \cdot \left(8 H - 10\right) \cdot H$
$A \left(H\right) = \left(4 H - 5\right) \cdot H$
$A \left(H\right) = 4 {H}^{2} - 5 H$

Area of the triangle when its height is 5 cm?
$A \left(5\right) = 4 {\left(5\right)}^{2} - 5 \left(5\right)$
$A \left(5\right) = 100 - 25 = 75$
$A \left(5 c m\right) = 75 c {m}^{2}$

Check:
Given that the base is $10 c m \text{ less than " (8*5cm) ", so it is } 30 c m$
Area calculation:
$\frac{1}{2} \cdot 30 c m \cdot 5 c m = 15 c m \cdot 5 c m = 75 c {m}^{2}$

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