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2 years ago

Can solve this problem please

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

3 0

Answer:

It is C

Step-by-step explanation:

C is true because you can make 12 per hour in the table shown

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What will slope intercept form be of the equation line 3x-4y=15

Klio2033 [76]

$the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\\\ 3x-4y=15 \ \ |\ subtract\ 3x\ to\ both\ sides\\\\-4y=-3x+15 \ \ | \ divide \ both \ sides\ by\ (-4) \\\\y= \frac{3}{4}x-\frac{15 }{4} \\\\slope : \ m=\frac{3}{4}$

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3 years ago

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How many tickets were sold between 12pm and 3pm?

andreyandreev [35.5K]

Answer:

approximately 670 maybe

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3 years ago

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The question is asked of the photo

aivan3 [116]

Pythagoras Theorem:
$\boxed { \boxed {a^2 + b^2 = c^2 \beta }}$

Apply Pythagoras Theorem:

a² + 6² = 11²

a² = 11² - 6²

a²= 85

a = √85

a = 9.22 (nearest hundredth)

Answer: RT = 9.22 cm

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3 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 20 ft from the path and is kept f

krek1111 [17]

Answer:

0.128rad/sec

Step-by-step explanation:

Let x represent the between the man and the point on the path

θ = the angle

dx/dt = 4 ft/s

dθ/dt = rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight

tan θ = x/20 ft

Cross Multiply

20tan θ = x

dx/dt = 20sec² θ dθ/dt

dθ/dt = 1/20 × cos² θ dx/dt

dθ/dt= 1/20 × cos² θ × 4

dθ/dt = 1/5 × cos² θ

Note : cos θ = 4/5

dθ/dt = 1/5 × (4/5)²

dθ/dt = 16/125

dθ/dt = 0.128rad/sec

6 0

3 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

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3 years ago

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