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2 years ago

Pls answer the 4 equations for the graph.

Mathematics

2 answers:

mihalych1998 [28]2 years ago

6 0

The second one is the correct one

AveGali [126]2 years ago

4 0

Number 1 is the -4x + 4 Number 2 is -1/2x -3
Number 3 is 4x-4 Number 4 is 1/3x - 4

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Plz help this one is the last one of the day plz help

Juliette [100K]

We know 20% is 15
and now we find 80%

since 20% x 4 = 80%, we can multiply 15 by 4 to get 80% of 75

15 x 4 = 60

80% of 75 = 60

3 0

2 years ago

12 is what percent of 19

VikaD [51]

Answer:

63%

Step-by-step explanation:

12/19=0.63157895

0.63157895*100=63.1578947

7 0

3 years ago

2x+3y=9<br> 2x+y=7<br> simultaneous equation

Marat540 [252]

Answer:

x=3 and y=1

Step-by-step explanation:

subtract equation 2 from 1

then (2x+3y) -(2x+y) =9-7

2x-2x+3y-y =2

2y=2 therefore y=1

substitute y value in 2x+y=7

2x+1=7

2x=7-1

2x=6

x=3

3 0

3 years ago

What is total number of elements in 3 trials of rolling a 6-sided die?

umka21 [38]

The answers should be 9

6 0

3 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago