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3 years ago

How can you solve problems by ordering rational numbers from least to greatest?

Mathematics

2 answers:

lina2011 [118]3 years ago

7 0

Answer:

B same form is the answer

Step-by-step explanation:

Mariana [72]3 years ago

6 0

Answer:

B same form

Step-by-step explanation:

By using them for fractions or decimals. You can compare fractions by having them from least to greatest and it helps in getting the same denominator for fractions.

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NEED ANSWER ASAP!!! (EASY)

Nata [24]

3000(0.25)=750

C is correct. Hope this helps!

4 0

3 years ago

Read 2 more answers

What is the height of the flagpole in the diagram below

Olegator [25]

Answer: The height of the flagpole is 12 feet.

Step-by-step explanation:

Let the height of the flagpole be 'H'.

So, from the figure, we can conclude that, the height and shadow of the flagpole is in proportion to the height and shadow of the stick.

Given:

Height of stick is,

Shadow of the stick is,

Shadow of the flagpole is,

Now, as these are in proportion, we have

Plug in the given values and solve for 'H'. This gives,

Hence, the height of the flagpole is 12 feet.

4 0

3 years ago

Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$