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d1i1m1o1n [39]
3 years ago
11

How many different triangles can you make if you are given these three lengths for sides 14,50,10

Mathematics
2 answers:
liberstina [14]3 years ago
8 0

Answer:

0

Step-by-step explanation:

arsen [322]3 years ago
4 0

Answer: 0 triangles

Step-by-step explanation: Speaking as if side a = 14, b = 60, and c = 10 -------

(a + c) > b

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Write the multiplicative inverse of <br>a) -13/19​
irga5000 [103]

Answer:

-  \frac{19}{13}

Step-by-step explanation:

the \: multiplicative \: inverse \: of \:  \\  -  \frac{13}{19}  =  -  \frac{19}{13}

Product of a number and its multiplicative inverse should be equal to 1.

So,

\bigg ( -  \frac{13}{19} \bigg)  \bigg ( -  \frac{19}{13} \bigg)

= \frac{13}{19}\times  \frac{19}{13}

=1

3 0
3 years ago
5x+2.5=-20 solve x, show your steps
notka56 [123]

Answer:

5x =  - 20 - 2.5 \\ 5x = −22.5 \\x =    \frac{ - 22.5}{ - 5}   \\ (x =  - 4.5) \times  - 1 \:  \:  \:  \: x \: can \: not \: be \: negative \\ x = 4.5

5 0
3 years ago
Chance has $5 in quarters. Blake has $8 in dollar coins.
artcher [175]
Blake because he has 8
3 0
3 years ago
How do the values compare? Order the values from least to greatest.
aleksley [76]
Im pretty sure it would be ordered: -2 1/2, -1/2, |1/2|, 1 1/2, |-2|, |-2 1/2|

This is because
-1/2 stays as -1/2
the absolute value of |-2| is 2
the absolute value of |1/2| is 1/2
the absolute value of |-2 1/2| is 2 1/2
-2 1/2 stays as -2 1/2
and 1 1/2 stays the same as well.
If you take all these numbers (-1/2, 2, 1/2, 2 1/2, -2 1/2, and 1 1/2) and order them from least to greatest, you would get:
-2 1/2, -1/2, 1/2, 1 1/2, 2, 2 1/2
which is -2 1/2, -1/2, |1/2|, 1 1/2, |-2|, |-2 1/2|

Im not exactly sure if it is all correct, but i hope this helps! :)
8 0
2 years ago
Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
kenny6666 [7]

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

  =\frac{-5+7}{4}

  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

3 0
4 years ago
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