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3 years ago

Look at pic for answer will mark brainilest PLEASE

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

5 0

Answer:

I believe B is the independent variable.

Step-by-step explanation:

an independent variable represents a quantity that can be manipulated in this situation it would the number of bushels that can be manipulated.

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A taxi travels for 30 minutes at a speed of 28 mph. Calculate how far it travles.

Pani-rosa [81]

Answer:

14 m

Step-by-step explanation:

Speed = 28 mph

Time = 30 minutes = 30/60 = 1/2 hour

Distance = speed * time

= 28 * $\frac{1}{2}$

= 14 m

3 0

3 years ago

Read 2 more answers

Point B is between A and C.<br> AB=4x - 6, BC=2x+2, and AC=44 in.<br> Find AB and BC.

GREYUIT [131]

Please double check my work
AB = -2x
BC = 4x

7 0

4 years ago

Write the equation of a line that passes through the given point and is parallel to the given line.

monitta

Answer:

y=2x+5
y= 1/3x -4

Step-by-step explanation:

$(1,-3)=(x_1,y_1)\\\\y =2x+2\\\\m = 2$

Substitute values into point-slope form

$y-y_1=m(x-x_1)\\\\y -3 =2(x-(-1))\\\\y-3=2(x+1)\\\\y-3=2x+2\\\\y=2x+2+3\\\\y=2x+5$

$(18,2) =(x_1,y_1)\\\\3y-x=-12$

Arrange equation in slope-intercept form (y=mx+b)and derive slope

$3y =x-12\\\\\frac{3y}{3} =\frac{1x}{3} -\frac{12}{3} \\\\y = \frac{1}{3} x -4\\\\m = \frac{1}{3}$

Substitute values into point-slope form

$y-y_1=m(x-x_1)\\\\\\y-2=\frac{1}{3} (x-18)\\\\y-2=\frac{1}{3} x -6\\\\y=\frac{1}{3} x -6+2\\\\y=\frac{1}{3} x -4$

5 0

3 years ago

At Southside middle school 40%of student play an instrument.there are 305 student.how many students play an instrument

Nadusha1986 [10]

You simply use this operation:
- total students are 305, we need to find how many of thm play instrumenst if 40% does.
40% can be rewrited as 0,4 in decimal.
than we simply utilise a basic percentage operation which allows us to find A PART of somethin=>
305×0,4= students who instruments.

8 0

4 years ago

Prove that :-<br><br> Cos3A-cosA/sin3A-sinA+cos2A-cos4A/sin4A-sin2A=sinA/cos2Acos3A

ad-work [718]

$\dfrac{\cos^3A-\cos A}{\sin^3A-\sin A}+\dfrac{\cos^2A-\cos^4A}{\sin^4A-\sin^2A}=\dfrac{\sin A}{\cos^2A\cos^3A}\\\\L_s=\dfrac{\cos A(\cos^2A-1)}{\sin A(\sin^2A-1)}+\dfrac{\cos^2A(1-\cos^2A)}{\sin^2A(\sin^2A-1)}\\\\=\dfrac{\cos A(-\sin^2A)}{\sin A(-\cos^2A)}+\dfrac{\cos^2A\sin^2A}{\sin^2A(-\cos^2A)}\\\\=\dfrac{\sin A}{\cos A}-1=\tan A-1$

$R_s=\dfrac{\sin A}{\cos A\cos^4A}=\dfrac{\sin A}{\cos A}\cdot\dfrac{1}{\cos^4A}=\tan A\cdot\dfrac{1}{\cos^4A}=\dfrac{\tan A}{\cos^4A}$

$\boxed{L_s\neq R_s}$

8 0

3 years ago