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3 years ago

5

Help!!!!!!!!! i am really confused and need help

1 answer:

stich3 [128]3 years ago

3 0

Answer:

g(-3)= -3^3-7

= -27-7

= -34

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. Two bicyclists, Sarah and Emily, depart from the same location. Sarah heads N55°E at an average speed of 6 km/hr. Emily heads

Liula [17]

Answer:

The distance between them after 30 minutes is 6.5 km.

Step-by-step explanation:

Speed = $\frac{distance}{time}$

Sarah's speed = 6 km/hr = 1.6667 m/s

Emily's speed = 10 km/hr = 2.7778 m/s

The measure of angle between their bearings = $105^{o}$

After 30 minutes (1800 seconds);

distance = speed x time

Sarah would have covered a distance = 1.6667 x 1800

= 3000 m

= 3 km

Emily would have covered a distance = 2.7778 x 1800

= 5000 m

= 5 km

The distance between them, a, can be determined by applying the cosine rule;

$a^{2}$ = $b^{2}$ + $c^{2}$ - 2bcCos A

= $5000^{2}$ + $3000^{2}$ -2(5000 x 3000) Cos 105

= $5000^{2}$ + $3000^{2}$ -2(5000 x 3000) x (-0.2588)

= 2.5 x $10^{7}$ + 9 x $10^{6}$ + 7764000

= 41764000

a = $\sqrt{41764000}$

= 6462.5073

a = 6462.5 m

The distance between them after 30 minutes is 6.5 km.

4 0

3 years ago

If f(x)=3x-6 then find f(4)

masya89 [10]

Answer:

f(4) = 6

Step-by-step explanation:

f(x) = 3x - 6

f(4) = 3(4) - 6

f(4) = 12 - 6

f(4) = 6

8 0

3 years ago

I really need help can somebody please tell me

kotegsom [21]

1-Like Terms
2-Unlike Terms
3-Like terms
4Like Terms
5-Unlike terms
6-Unlike Terms

7 0

3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

4 years ago

Read 2 more answers

ʄɨռɖ tɦɛ ʋaʟʊɛ օʄ 7x? <br>2

LUCKY_DIMON [66]

Answer:

7x when x=2=<em>7</em><em>×</em><em>2</em><em>=</em><em>1</em><em>4</em><em> </em><em>is</em><em> </em><em>your</em><em> </em><em>answer</em>

4 0

3 years ago