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pickupchik [31]

2 years ago

5

Lucia compro 12 pares de media a 60 pesos cada una 7 remeras a 156 pesos cada una y 2 pantalones a 350 pesos cada uno.Si pago co

n 3000 pesos¿le sobro o le falto dinero?¿cuanto?

1 answer:

earnstyle [38]2 years ago

5 0

Answer:

A Lucia le sobró dinero y la cantidad que sobró es: 488 pesos.

Step-by-step explanation:

Para determinar si le faltó o le sobro dinero, debes determinar el costo total de lo que compró multiplicando el precio de cada artículo por la cantidad comprada y sumando estos valores:

(60*12)+(156*7)+(350*2)=720+1092+700=2512

Dado que pagó con 3000 pesos y el costo total de lo que compró es de 2512 pesos y este valor es menor que 3000 en 488 pesos, la respuesta es que le sobró dinero.

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What is the value of this expression 32+50+(-50) HELP ME IM BEGGING YOUUUU

skad [1K]

Answer:-
So if ur question merely includes addition and subtraction operations then it’s “32”

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2 years ago

olya-2409 [2.1K]

Echa t-shirt is 6 dollars and ecah hat is 5
29-23=6
And 6•4=24
29-24=5

4 0

2 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

2 years ago

Read 2 more answers

Which of the following relationships is a direct variation?

mafiozo [28]

Answer:Well All i know is that the answer is not C I'm sorry I dont know the correct answer but I know for a fact that the answer isn't C

Step-by-step explanation:

4 0

3 years ago

If cos theta= -8/17 and theta is in quadrant 3, what is cos2 theta and tan2 theta

Karo-lina-s [1.5K]

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad 
\begin{array}{llll}
\textit{now, hypotenuse is always positive}\\
\textit{since it's just the radius}
\end{array}
\\\\\\
thus\qquad cos(\theta)=\cfrac{-8}{17}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}$

since the hypotenuse is just the radius unit, is never negative, so the - in front of 8/17 is likely the numerator's, or the adjacent's side

now, let us use the pythagorean theorem, to find the opposite side, or "b"

$\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\\\\\
\pm\sqrt{17^2-(-8)^2}=b\implies \pm\sqrt{225}=b\implies \pm 15=b$

so... which is it then? +15 or -15? since the root gives us both, well
angle θ, we know is on the 3rd quadrant, on the 3rd quadrant, both, the adjacent(x) and the opposite(y) sides are negative, that means, -15 = b

so, now we know, a = -8, b = -15, and c = 17
let us plug those fellows in the double-angle identities then

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\
cos(2\theta)=1-2sin^2(\theta)\implies cos(2\theta)=1-2\left( \cfrac{-15}{17} \right)^2
\\\\\\
cos(2\theta)=1-\cfrac{450}{289}\implies cos(2\theta)=-\cfrac{161}{289}$

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\left( \frac{-15}{-8} \right)}{1-\left( \frac{-15}{-8} \right)^2}
\\\\\\
tan(2\theta)=\cfrac{\frac{15}{4}}{1-\frac{225}{64}}\implies tan(2\theta)=\cfrac{\frac{15}{4}}{-\frac{161}{64}}
\\\\\\
tan(2\theta)=\cfrac{15}{4}\cdot \cfrac{-64}{161}\implies tan(2\theta)=-\cfrac{240}{161}$

6 0

3 years ago