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Helga [31]
3 years ago
12

Number 14 please help

Mathematics
2 answers:
Natasha_Volkova [10]3 years ago
7 0
Hey!! the answer to the problem is 15680!you get it by multiplying the 3 numbers!
hope i helped!! :))
liberstina [14]3 years ago
4 0
I think you should say you multipliyed?
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Test question what would the sign be when multiplying a positive and negative number
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2 years ago
It is known that 1.4< square root of 2 <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
The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad &#10;C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}&#10;\\\\\\&#10;AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\&#10;-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad &#10;D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}&#10;\\\\\\&#10;BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}&#10;\\\\\\&#10;BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]&#10;\\\\\\&#10;4[15]\implies 60

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3 years ago
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