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3 years ago

The line y = 7x - 2 has a slope of 7 and a y-intercept of (-2 ,0). TRUE OR FALSE.

Mathematics

1 answer:

8_murik_8 [283]3 years ago

5 0

The y int. is correct but you wrote it wrong it is (0, -2)

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1. Given line segment RB and on it is point C, if RB = 45 and CB = 3x - 12 and RC = 2x + 9, what is the value of x?

marshall27 [118]

Answer:

Step-by-step explanation:

1). If a point C is lying on a line segment RB,

Then RB = RC + CB

Since, RB = 45, CB = 3x - 12 and RC = 2x + 9

45 = (2x + 9) + (3x - 12)

45 = (2x + 3x) - 3

48 = 5x

x = $\frac{48}{5}$

x = 9.6

2). Since, RC = 2x + 9

RC = 2(9.6) + 9

= 19.2 + 9

= 28.2 units

3). Distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between the points A(2, -2) and D(-2, 2) will be,

$AD=\sqrt{(2+2)^2+(-2-2)^2}$

= $\sqrt{16+16}$

= $4\sqrt{2}$

= 5.657

≈ 5.66 units

7 0

3 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72