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rjkz [21]
3 years ago
7

What is the answer ?????​

Mathematics
1 answer:
AfilCa [17]3 years ago
8 0

Answer:

2456≈2500

14725=14700 (to 3sf)

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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}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

4 0
2 years ago
5k+ 3 &lt;-32 or -4k+ 7 &lt;-29
inessss [21]

Answer:

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
It is known that 1.4&lt; square root of 2 &lt;1.5. Find all possible values of the expression 2-square root of 2. How do I find
horsena [70]

Answer:

0.5<2-√2<0.6

Step-by-step explanation:

The original inequality states that 1.4<√2<1.5

For the second inequality, you can think of 2-√2 as 2+(-√2).

Because of the "properties of inequalities", we know that when a positive inequality is being turned into a negative, the numbers need to swap and become negative. So, the original inequality becomes -1.5<-√2<-1.4. (Notice how the √2 becomes negative, too). This makes sense because -1.5 is less than -1.4.

Using our new inequality, we can solve the problem. Instead of 2+(-√2), we are going to switch "-√2" with both possibilities of -1.5 and -1.6. For -1.5, we would get 2+(-1.5), or 0.5. For -1.4, we would get 2+(-1.4), or 0.6.

Now, we insert the new numbers into the equation _<2-√2<_. The 0.5 would take the original equation's "1.4" place, and 0.6 would take 1.5's. In the end, you'd get 0.5<2-√2<0.6. All possible values of 2-√2 would be between 0.5 and 0.6.

Hope this helped!

4 0
3 years ago
3.305 + 1.7 what's the sum? Also more questions are coming just doing a test
Paraphin [41]

  3.305

 +1.7

  5.005

Line up the decimal points and add like you usually would add.

5.005 is the answer.

5 0
3 years ago
LOL ANOTHER MATH QUESTION ASAP!!!
Inessa05 [86]
Your answer is the first one q
6 0
3 years ago
Read 2 more answers
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