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Tanzania [10]
3 years ago
14

What is the approximate solution to this equation?

Mathematics
2 answers:
Sindrei [870]3 years ago
5 0

Answer: b)

Step-by-step explanation:

3^3 is equal to 27 exactly, but it asks for an approximate. 27.55 is near 27 so yea, there ya go ^

ankoles [38]3 years ago
4 0
B: the cube root of 27.55 is rounded to 3

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bulgar [2K]
B, divide it’s coordinates in half. -5/2 and -1/2.
3 0
2 years ago
The temperature rose from -7°F to 21°F. By how much did the temperature increase?
horrorfan [7]

Answer: 28

Step-by-step explanation: you have to add 21 + 7 because your counting up from -7 to 21

7 0
3 years ago
Graph to answers i put
maksim [4K]

Answer:

The domain of f is {x : x ∈ R, -7 ≤ x ≤ 7} ⇒ [-7, 7]

Step-by-step explanation:

<em>The domain of a function is </em><em>all the values of x</em><em> make the function defined</em>

In the given graph

∵ The graph drawn from x = -7 to x = 7 and from y = -6 to y = 4

→ That means the values of x started from -7 and ended at x = 7

∵ The coordinates of the starting point of the graph of f are (-7, -5)

∵ The coordinates of the ending point of the graph of f are (7, 3)

∴ -7 ≤ x ≤ 7

∵ Values of x are the domain of the function

∴ The domain of the function is -7 ≤ x ≤ 7

∴ The domain of f = {x : x ∈ R, -7 ≤ x ≤ 7} ⇒ [-7, 7]

5 0
3 years ago
I don’t know how to find angle A
Cerrena [4.2K]
Angle a = 60

given that all angles of a triangle added together must equal 180
x = -20
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60+60+40=180

therefore, angle A is 60°
5 0
3 years ago
Use Lagrange multipliers to find the maximum and minimum values of
marusya05 [52]

The Lagrangian is

L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)

It has critical points where the first order derivatives vanish:

L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}

L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y

L_\lambda=x^2+y^2-4=0

From the first two equations we get

-\dfrac1{2x}=-\dfrac4y\implies y=8x

Then

x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}

At these critical points, we have

f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125 (maximum)

f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125 (minimum)

5 0
3 years ago
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