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3 years ago

The function f(x) is shown on the graph.

Mathematics

2 answers:

grandymaker [24]3 years ago

8 0

Answer:

option A

$12$ only

Step-by-step explanation:

we know that

The y-intercept of a function is the value of the function when the value of x is equal to zero

In this problem

f(0) represent the value of the function when x is equal to zero

therefore

f(0) is the y-intercept of the function

see the graph

for $x=0$

$f(0)=12$

enyata [817]3 years ago

6 0

Answer A i think should be right

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The unit rate is 2.5 pounds a week.
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tresset_1 [31]

The perimeter means the sum of all sides.
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-2/3-5/6= write in simplest form

sleet_krkn [62]

-2/3 - 5/6

We need common denominators so, since 3 can go into 6, we only need to multiply the first fraction by 2.

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3 years ago

Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a

Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force ( $\vec F$ ), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

$\vec F = \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right)$ (1)

Vector (ordered pair - polar form)

$\vec F = \left(\|\vec F\|, \theta\right)$

Sum of vectorial components (linear combination)

$\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j}$ (2)

Where:

$\|\vec F\|$ - Norm of the vector force, in newtons.

$\theta$ - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

$\hat{i}$ , $\hat{j}$ - Orthogonal axes, no unit.

If we know that $\|\vec F\| = 30\,lbf$ and $\theta = 25^{\circ}$ , then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

$\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]$

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]$

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

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3 years ago

1) What is the solution of the given system?

m_a_m_a [10]

Answer:

1 x=-2.5 y = -5.5

2. x=5 y=1

Step-by-step explanation:

1) What is the solution of the given system?

5x-y=-7

3x-y=-2

Multiply the second equation by -1

-1*(3x-y)=-1(-2)

-3x +y = 2

Now add the first equation to the modified second equation

5x-y=-7

-3x +y = 2

------------------

2x = -5

Divide each side by 2

2x/2 = -5/2

x = -2.5

Now we need to find y

-3x+y =2

-3(-2.5) +y =2

7.5 +y =2

Subtract 7.5 from each side

7.5 -7.5 +y =2-7.5

y = -5.5

2) what is the solution of the given system?

5x+7y=32

8x+6y=46

Divide the second equation by 2

8x/2+6y/2=46/2

4x+3y =23

Multiply the first equation by 4

4 (5x+7y)=32*4

20x+28y = 128

Now multiply the modified 2nd equation by -5

-5(4x+3y )=-5(23 )

-20x -15y = -115

Lets add the new equations together to eliminate x

20x+28y = 128

-20x -15y = -115

---------------------

13y = 13

Divide each side by 13

13y/13 =13/13

y=1

Now substitute back in to find x

5x+7y=32

5x +7(1) =32

5x +7 =32

Subtract 7 from each side

5x+7-7 =32-7

5x =25

Divide by 5

5x/5 =25/5

x=5

3 0

4 years ago