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4 years ago

What I'd the square root of 81

Mathematics

2 answers:

Ludmilka [50]4 years ago

5 0

the square root of 81 is 9

omeli [17]4 years ago

3 0

The square root of 81 is 9 since 9 multiplied by itself is 81 (using the power of two)

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What is the slope of the line that passes through the points (−10,−4) and (10, -29)?

OleMash [197]

Answer:

slope= - 5/4

Step-by-step explanation:

3 0

3 years ago

in a new clothing store, she charges 21.75 for each sweatshirt she sells. what was the revenue if she sold 51 sweatshirts last m

finlep [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

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Ecuación de la hipérbola con centro en (0;0), focos en abrir paréntesis 0 coma espacio menos raíz cuadrada de 28 cerrar paréntes

yaroslaw [1]

Answer:

$\frac{y^{2}}{25}-\frac{x^{2}}{3}=1$

Step-by-step explanation:

Para resolver este problema debemos tomar en cuenta los datos que nos dan y la ecuación de una hipérbola. Comencemos con los datos:

centro: (0,0)

focos: $(0,-\sqrt{28}),(0,\sqrt{28})$

eje conjugado = $2\sqrt{3}$

por los focos podemos ver que la hipérbola se dirige hacia el eje y, por lo que debemos tomar la siguiente forma de la ecuación de la parábola:

$\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1$

de los focos podemos obtener que:

$c=\sqrt{28}$

y del eje conjugado podemos saber que al dividir la longitud del eje conjugado dentro de 2 obtenemos b, así que:

$b=\sqrt{3}$

podemos utilizar la siguiente fórmula para obtener a:

$c^{2}-a^{2}=b^{2}$

si despejamos a en la ecuación obtenemos lo siguiente:

$a=\sqrt{c^{2}-b^{2}}$

ahora podemos sustituir los valores:

$a=\sqrt{(\sqrt{28})^{2}-(\sqrt{3})^{2}}$

$a=\sqrt{28-3}$

$a=\sqrt{25}$

a=5

así que media vez conozcamos a, podemos sustituir los datos en la ecuación de la hipérbola así que obtenemos lo siguiente:

$\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1$

$\frac{y^{2}}{(5)^{2}}+\frac{x^{2}}{(\sqrt{3})^{2}}=1$

$\frac{y^{2}}{25}+\frac{x^{2}}{3}=1$

si graficamos la hipérbola, queda como en el documento adjunto.

7 0

3 years ago

Proof that :<br><img src="https://tex.z-dn.net/?f=%20%7Bsin%7D%5E%7B2%7D%20%5Ctheta%20%2B%20%20%7Bcos%7D%5E%7B2%7D%20%5C%3A%5Cth

Arisa [49]

Answer:

Solution given:

Right angled triangle ABC is drawn where <C= $\theta$

we know that

$\displaystyle Sin\theta=\frac{opposite}{hypotenuse} =\frac{AB}{AC}$

$\displaystyle Cos\theta=\frac{adjacent}{hypotenuse}=\frac{BC}{AC}$

Now

left hand side

$\displaystyle {sin}^{2} \theta + {cos}^{2} \:\theta$

Substituting value

$(\frac{AB}{AC})²+(\frac{BC}{AC})²$

distributing power

$\frac{AB²}{AC²}+\frac{BC²}{AC²}$

Taking L.C.M

$\displaystyle \frac{AB²+BC²}{AC²}$ ....[I]

In ∆ABC By using Pythagoras law we get

$\boxed{\green{\bold{Opposite²+adjacent²=hypotenuse²}}}$

AB²+BC²=AC²

Substituting value of AB²+BC² in equation [I]

we get

$\displaystyle \frac{AC²}{AC²}$

Right hand side

<h3><u>proved</u></h3>

8 0

3 years ago

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A biologist found a system of equations to help her calculate the growth of bacteria. The functions she found y=2x2−15y=2x2-15 &

Nesterboy [21]

Part A.
The set of equations can be solved by substitution. Use the expression one equation gives for y as the value of y in the other equation. This gives
2x²-15 = 3x-6
Subtracting the right side gives a quadratic in standard form that can be solved by any of several methods.
2x² -3x -9 = 0
(2x+3)(x-3) = 0 . . . . factor the above equation
x = -3/2, x = 3 . . . . .use the zero product rule to find x
Now, these x-values can be substituted into either equation for y. The linear equation is often easier to evaluate.
y = 3(-3/2) -6 = -10.5
y = 3(3)-6 = 3

The solutions to the system are (-1.5, -10.5) and (3, 3).

Part B.
The two equations can be graphed. The solutions are where the graphs intersect. The graphs intersect where the (x, y) values that satisfy one equation are the same (x, y) values that satisfy the other equation. Those points of intersection are (-1.5, -10.5) and (3, 3).

6 0

3 years ago