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marissa [1.9K]
3 years ago
15

$30 per hour it cut off pls help i give brainlest if it’s correct

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

t  = 4 h

Step-by-step explanation:

Painter A   initial fee 40 $  and each hour charge 45 $

Painter B   initial fee 100 $ and each hour charge  30 $

The equations for painters are straight lines.

The initial fee is the intercept of the line with the y-axis ( painters receive 40 and 100 $ respectively and they have done nothing.

The equation for a straight line is:

y = m*x + b       where m is the slope and b the intercept with the y-axis

Then

Painter A     when x = 0  y = b = 40

And  each hour of work cost 45 $   then f  x = t  in hours

The equation for Painter A is   y  = 45*t  + 40

Similarly, the equation for Painter B is

y  = 30*t + 100

We have a two equations system, the solution will indicate when the two painters will charge the same amount of money

y  =  45*t  + 40      (1)

y  =  30*t  + 100     (2)

Equation 1 - equation 2

0  =  15*t  - 60

15*t  =  60

t  = 60/15

t  = 4 h

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A line passing through which of the following pairs of coordinates represents a proportional relationship? A. (1, 3) and (2, 6)
trasher [3.6K]

Answer:

Answer is A - (1, 3) and (2, 6)

Step-by-step explanation:

(1, 3) • 2

= (1 • 2, 3 • 2)

= (2, 6)

Ratio: 2/2, 6/2

Therefore, the answer is A.  

3 0
3 years ago
Read 2 more answers
Help pleassssee
maks197457 [2]
150 * 1.25 * 1.25* 1.25 = £292.97
Hope it helps
6 0
3 years ago
Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you
Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi, we should expect \cos x. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi

so we should expect \tan\dfrac x2 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

3 0
3 years ago
Task 4:
Zarrin [17]

B i think i am sorry if it is not right but i am pretty sure i am right tho

6 0
3 years ago
Which transformations could be performed to show that
emmainna [20.7K]

Answer:

no picture given

Step-by-step explanation:

no picture given, so it is impossible to say.

3 0
3 years ago
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