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xeze [42]
2 years ago
6

A law firm charges a client according to the function f(x)=4(x+2) , where x represents the number of hours spent on the client’s

case each week, and f(x) represents the total charge that week in hundreds of dollars. Which graph best represents f(x)? A
image 9aa1ffbf4b6c4b819528d03e38d40220

B
image 2613910ac77a443ab121eae0681d4242

C
image 9a1a98d3a68b4a5b91b8831ed35dc6ef

D
image 85146317251e4058afc4ecb02264fffa

Mathematics
1 answer:
Zigmanuir [339]2 years ago
3 0

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x ---> the number of hours spent on the client’s case each week

f(x) ---->  the total charge that week in hundreds of dollars

we have

f(x)=4(x+2)

distribute

f(x)=4x+8

This is the equation of the line in slope intercept form

where

The slope or unit rate is

m=4

The y-intercept or initial value is

b=8

To graph the line we need minimum two points

Find the x-intercept

For f(x)=0

0=4x+8\\4x=-8\\x=-2

The x-intercept is the point (-2,0)

so

we have

the points (-2,0) and (0,8)

To graph the line, plot the intercepts, connect them and join the points

The graph in the attached figure

Remember that the value of x and the value of f(x) cannot be a negative number

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M = 331,482 for Brooklyn.

Which means Brooklyn's method will pay of more.
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In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
Sergeeva-Olga [200]

Answer:

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0
2 years ago
Which expression could be used to determine the product of –4 and 3 and one - fourth?
ELEN [110]

The correct answer is (-4)(3)+(-4)(1/4)

3 0
2 years ago
In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a
geniusboy [140]

Answer:

BC:BN=8:3

Step-by-step explanation:

ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.

Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD

But, BC= BN+NC

Therefore, BC:BN=(BN+NC):BN

⇒BC:BN=(3+5):3

⇒BC:BN=8:3


4 0
2 years ago
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