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Lostsunrise [7]
3 years ago
11

The y-intercept is the point where the graph crosses the __ axis

Mathematics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

what are the Options?

Step-by-step explanation:


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What's a real number between -1/10 and 0?
vova2212 [387]

Answer:

The answer is at the bottom!!

Step-by-step explanation:

And All of those Natural Numbers will be starting with 1 so we will be having other Numbers as well on which No Number will be mapped like 2 or 211 or 79 So This Means Set of Natural Numbers is Grater then Real Numbers Between 0 and 1. So Set of Real Numbers Between 0 and 1 is Countably Infinite.

Hope this helps!!

4 0
3 years ago
Hola esta es mi pregunta
fenix001 [56]

Answer:

Step-by-step explanation:

hola

Cuando dos numeros tienen el mismo signo la respuesta siempre es positiva. cuando un signo es positivo y e lotro es negativo la respuesta siempre sera negativa. Esta es la regla para multiplicar numeros con signos diferentes. el orden de los signos no importa en multiplicacon siempre va ha ser la misma respuesta.

7 0
3 years ago
If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,
S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
3 0
3 years ago
Will mark Brainliest!
Anettt [7]

Answer:a=2.1 :)

Step-by-step explanation:a+(-1.6)=-3.7

+1.6 +1.6

a=-2.1

5 0
3 years ago
Show that the equation x^3 + x -3 =0 has a solution between 1 and 2. That is the only information given
valentina_108 [34]

Answer:

Step-by-step explanation:

Given that:

f(x) = x^3 + x - 3 = 0

Since the given function is equal to zero, then the function :

f(x) = x^3 + x - 3

where;

x = 1 and

x = 2

f(x) = (1)^3 + (1) - 3 \\  \\ f(x) = 1 + 1 - 3 \\ \\f(x) = -1 < 0

f(2) =  x^3 + x - 3 \\ \\f(2) = (2)^3 + 3 - 3  \\ \\f(2) = 8 + 3 - 3  \\ \\f(2) = 11 - 3 \\ \\f(2) = 8 > 0 \\ \\

Thus; f(1) < 0 and f(2) > 0

8 0
3 years ago
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