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3 years ago

Please answer quickly. I need help bad

Mathematics

1 answer:

weeeeeb [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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(Q5) Decide if the function is an exponential function. If it is, state the initial value and the base. y=3.1^x

ycow [4]

Answer:

Step-by-step explanation:

Exponential equation takes the form $y=ab^x$ where

a is the initial value ( a ≠ 0), and

b is the base ( b ≠ 1)

The equation given can be written as $y=3.1^x\\y=1*3.1^x$ . Thus, it is an exponential equation.

So a = 1 and b = 3.1

Thus we can say that the initial value = 1 and the base is 3.3

6 0

3 years ago

Properties of square

nikklg [1K]

a square is a quadrilateral

5 0

4 years ago

Find the exact length of the curve. x = 1 + 6t2, y = 4 + 4t3, 0 ≤ t ≤ 3

vovangra [49]

$x=1+6t^2\implies\dfrac{\mathrm dx}{\mathrm dt}=12t$

$y=4+4t^3\implies\dfrac{\mathrm dy}{\mathrm dt}=12t^2$

The length of the curve $C$ is given by

$\displaystyle\int_C\mathrm ds=\int_0^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$=\displaystyle\int_0^3\sqrt{144t^2+144t^4}\,\mathrm dt$

$=\displaystyle12\int_0^3t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3 2t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3\sqrt{1+t^2}\,\mathrm d(1+t^2)$

$=6\cdot\dfrac23(1+t^2)^{3/2}\bigg|_0^3$

$=4(10^{3/2}-1)$

$=40\sqrt{10}-4$

8 0

3 years ago

An engineer want to fill a 2m height cylindrical with concrete if the radius of concrete is0.5m, how much of volumn of concrete

lina2011 [118]

Answer:

v=20h 2(314)(o.5m) 2m= 6.28m3

Step-by-step explanation:

3 0

2 years ago

Determine whether each point lies on the graph of the equation.(a) 2x - y - 3 = 0 (a) (1, 2)Yes, the point is on the graph.No, t

Neko [114]

Answer:

a) (1, 2) not on the graph.

b) (1, -1) is on the graph.

Step-by-step explanation:

Given the equation of the line as:

$2x - y - 3 = 0$

The points given are

a) (1, 2)

To determine whether it is on the graph of the line or not.

To do so, we can do 2 things:

1. Draw the graph and plot the point on the graph to check whether it is on the graph or not.

2. To put the point in the given equation of the line, whether the equation is satisfied or not.

For method 1: Kindly refer to the attached image of the line and point plotted.

Method 2:

Let us put $x=1, y=2$ in the Left Hand Side (LHS) of equation.

$2\times 1 - 2 - 3\\\Rightarrow -3 \neq 0 , RHS$

(1, 2) Not on the graph.

(b) (1, -1)

For method 1: Kindly refer to the attached image of the line and point plotted.

Method 2:

Let us put $x=1, y=-1$ in the Left Hand Side (LHS) of equation.

$2\times 1 - (-1) - 3\\\Rightarrow 3-3 = 0 = RHS$

(1, -1) is on the graph.

3 0

4 years ago