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11Alexandr11 [23.1K]
3 years ago
5

Help me with this two questions

Mathematics
1 answer:
Anni [7]3 years ago
5 0
First question: 0

Second question: 0.94
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HELP PLEASE !?!? I suck at math
Anna35 [415]
To graph the equation, plot a few points then connect then with a line.

f(x) = 440 - 55x

Plug in numbers for x to get the y value.

f(0) = 440
f(1) = 385
f(2) = 330
f(3) = 275
f(4) = 220
f(5) = 165

So when you graph these points, the y intercept is y = 440 because the graph crosses the y axis at (0, 440).

The x intercept is where the graph crosses the x axis. This happens when y = 0.

0 = 440 - 55x. Solve for x.

55x = 440
x = 8

So the x intercept is x = 8




7 0
3 years ago
Can someone help me?
mestny [16]

Answer:

Step-by-step explanation:

To do this, we must use the unit circle. At the angle \frac{\pi }{6} the coordinate pair is (\frac{\sqrt{3} }{2}, \frac{1}{2}) and the hypotenuse is 1

Now that we know these we can start using the trig functions

sin = y

cos = x

tan = \frac{y}{x}

And the other functions would be the reciprocal of these

csc = \frac{1}{y}

sec = \frac{1}{x}

cot = \frac{x}{y}

This would mean that

sinx = \frac{1}{2}

cosx =\frac{\sqrt{3} }{2}

tanx =\frac{1}{\sqrt{3} } or \frac{\sqrt{3} }{3} if you're supposed to rationalize your denominators

cscx = 2

secx = \frac{2}{\sqrt{3} } or \frac{2\sqrt{3} }{3}

cot = \sqrt{3}

6 0
3 years ago
Can someone help me on this math problem PLEASE
egoroff_w [7]

Answer:

QR perpendicular to PT

PR^Q=SR^T=90°

also ANGLES (QPR= STR

7 0
3 years ago
Lcm of 63, 80 and 102
photoshop1234 [79]

Answer:

85,680

Step-by-step explanation:

3 0
2 years ago
C=-7/15 and x=2/5 c-4x write your answer in simplest form
pogonyaev

9514 1404 393

Answer:

  -31/15

Step-by-step explanation:

Put the numbers in the expression and do the arithmetic.

  c-4x=-\dfrac{7}{15}-4\cdot\dfrac{2}{5}=-\left(\dfrac{7}{15}+\dfrac{8}{5}\right)\\\\=-\left(\dfrac{7}{15}+\dfrac{24}{15}\right)=\boxed{-\dfrac{31}{15}}

__

Your graphing calculator can probably help with this.

5 0
3 years ago
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