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3 years ago

Solve for x. Geometry

Mathematics

2 answers:

Travka [436]3 years ago

7 0

Answer:

Step-by-step explanation:

Side a = 17

Side b = 8

Side c = 15

Angle ∠A = 90° = 1.5708 rad = π/2

Angle ∠B = 28.072° = 28°4'21" = 0.48996 rad

Angle ∠C = 61.928° = 61°55'39" = 1.08084 rad

C=61.928°B=28.072°A=90°b=8a=17c=15

Area = 60

Perimeter p = 40

Semiperimeter s = 20

Height ha = 7.05882

Height hb = 15

Height hc = 8

Median ma = 8.5

Median mb = 15.52417

Median mc = 10.96586

Inradius r = 3

Circumradius R = 8.5

Vertex coordinates: A[0, 0] B[15, 0] C[0, 8]

Centroid: [5, 2.66667]

Inscribed Circle Center: [3, 3]

Circumscribed Circle Center: [7.5, 4]

Korvikt [17]3 years ago

5 0

Maybe the second one is correct

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She would sold 255 cups.

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2 years ago

The product of the number and itself is 361

mafiozo [28]

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4 years ago

Is 89/100 terminating or repeating

Margarita [4]

Answer:

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Step-by-step explanation:

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3 years ago

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There are some Rs2 and Rs5 coins in a box. The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3. The value of

Mars2501 [29]

Given:

The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.

The value of all the Rs5 coins is Rs45.

To find:

The value of all the Rs2 coins in the box.

Solution:

Let x be the number of Rs2 coins and y be the number of Rs5 coins.

The value of all the Rs5 coins is Rs45.

$5y=45$

$y=\dfrac{45}{5}$

$y=9$

The ratio of the number of Rs2 coins to the number of Rs5 coins is 1:3.

$\dfrac{x}{y}=\dfrac{1}{3}$

$\dfrac{x}{9}=\dfrac{1}{3}$

Multiply both sides by 9.

$\dfrac{x}{9}\times 9=\dfrac{1}{3}\times 9$

$x=3$

The value of all the Rs2 coins in the box is:

$\text{Required value}=2x$

$\text{Required value}=2(3)$

$\text{Required value}=6$

Therefore, the value of all the Rs. 2 coins in the box is Rs. 6.

4 0

3 years ago

Determinar el precio del artículo y la cantidad que debe producirse para obtener la máxima utilidad.?

Xelga [282]

Answer:

Se deben producir 15 artículos a un precio de 55 cada uno

Step-by-step explanation:

Parece que hay un error en el planteamiento del problema porque se obtienen ganancias negativas en cualquier solución óptima. Sin embargo, te mostraré el procedimiento correcto necesario.

La ecuación de la demanda es

$p = 100-3x$

El ingreso es el producto del precio por el número de artículos producidos

$I(x)=px=100x-3x^2$

El costo de producir x unidades viene dado por

$C= \frac{1}{2}x^2 -3x + 1500$

La Ganancia es la diferencia entre el ingreso y los costos

$G(x)=I-C=100x-3x^2-\left ( \frac{1}{2}x^2 -3x + 1500 \right )$

Operando

$G(x)=-\frac{7}{2}x^2+103x- 1500$

Para optimizar la ganancia, se obtiene su primera derivada y se iguala a cero

$G'(x)=-7x+103$

$-7x+103=0$

Obtenemos así el valor óptimo de producción

$x=\frac{103}{7}=14.7$

Siendo x un número entero por ser el número de unidades, probamos con los enteros que rodean a la respuesta obtenida

$G(14)=-744$

$G(15)=-742.5$

La menos negativa ganancia sería el valor óptimo, que según mencionamos, es negativa. Luego el valor óptimo es x=15

El precio con x=15 es

$p = 100-3(15)=55$

5 0

4 years ago