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pav-90 [236]
3 years ago
7

1/3 + 5/6 + 5/12 What is the answer

Mathematics
2 answers:
AlladinOne [14]3 years ago
6 0
Set all of the denominators to 12 by multiplying 1/3 by 4/4, and 5/6 by 2/2. Now just add them together. 4/12+10/12+5/12=19/12. Or 1 and 7 twelves.
SOVA2 [1]3 years ago
3 0
1/3 + 5/6 + 5/12

1.583333333333333
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452 x 12 _____ = 452<br> fill in the blank
Aloiza [94]

Answer:

38 remander 8

Step-by-step explanation:

5 0
3 years ago
What is the solution set of {x | x &gt; -3} {x | x &lt; 3}?
givi [52]

x > -3

x < 3

<em><u>-3 < x < 3</u></em>

question didn't mention about "x is real number, or positive, or negative etc".. so we can't say anything more

for example: x can be -2,-1,0,1,2

but if only positive, x can be = 1,2

" " negative, x can be = -1, -2

Hope this helps ^-^

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3 years ago
Determine if the statement below is always, sometimes, or never true. There are 250 degrees in the sum of the interior angles of
ZanzabumX [31]
It would never be true.
5 0
3 years ago
Find the domain of the function.
timurjin [86]

When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.

u(x) ≥ 0.

\sqrt{9x+27} \geq  0

9x + 27 ≥ 0 by squaring both sides

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So the domain of the function is when x ≥ -3 is true.

5 0
4 years ago
Tienes un triángulo equilátero con lados de 10cm, lo divides en dos partes de manera horizontal tal que el área de ambos figuras
Tanya [424]

Answer:

x = 10/√2 ≈ 7.07

Step-by-step explanation:

Comenzaremos por dividir el triángulo en dos partes y definir H, como en la figura adjunta.

Aplicando el teorema de Tales, sabemos que:

\dfrac{l/2}{H}=\dfrac{x/2}{h}\\\\\\\dfrac{l}{H}=\dfrac{x}{h}

También sabemos que, dado que el tirángulo menor es la mitad que el triángulo mayor, la relación entre áreas es:

\dfrac{A}{A_x}=\dfrac{lH/2}{xh/2}=\dfrac{lH}{xh}=2

Dado que formamos dos triángulos rectángulos, podemos despejar el valor de H como:

(l/2)^2+H^2=l^2\\\\H^2=l^2-(l/2)^2=10^2-5^2=100-25=75\\\\H^2=\sqrt{75}

Podemos entonces despejar x de la siguiente manera:

h=\dfrac{H}{l}\cdot x=\dfrac{lH}{2x}\\\\\\2x^2\dfrac{H}{l}=lH\\\\\\2x^2=l^2\\\\\\x=\dfrac{l}{\sqrt{2}}=\dfrac{10}{\sqrt{2}}\approx7.07

6 0
3 years ago
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