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3 years ago

Write the equations of 2 straight lines which are perpendicular to y = -3x +0.5.

Mathematics

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

y = (1/3)x + 7

y = (1/3)x - 15

These are two examples out of an unlimited number of possibilities.

Step-by-step explanation:

The slope of this given line is -3. Any new line perpendicular to this one is the negative reciprocal of -3: 1/3.

The slope-intercept form is y = mx + b. Here m = 1/3, so the revised form is

y = (1/3)x + b.

Two straight lines perpendicular to the given line are easily found by selecting two different values for b:

y = (1/3)x + 7

y = (1/3)x - 15

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2 years ago

Find the value of f(-6).

Karo-lina-s [1.5K]

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3 years ago

What is 0.000 002 in standard form

Vsevolod [243]

Answer:

$0.000 002 = 2 \times {10}^{ - 6}$

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

Log 10(x + 3) – log 10(x-3) = 1

nadezda [96]

Answer:33/9

Step-by-step explanation:

Log10(x+3)-Log10(x-3)=1

Log10((x+3)/(x-3))=1

(x+3)/(x-3)=10^1

(x+3)/(x-3)=10

Cross multiply

x+3=10(x-3)

Open brackets

x+3=10x-30

Collect like terms

10x-x=30+3

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4 0

4 years ago

Precalculus PLEASE HELP ME right answers only check all please

Alex_Xolod [135]

Given :

$cos(x)cos(\frac{\pi }{7}) + sin(x)sin(\frac{\pi }{7})=-\frac{\sqrt{2}}{2}$

We know the identity

$cos(x)cos(y) + sin(x)sin(y)=cos(x-y)$

We use this property to simplify the left hand side

So $cos(x)cos(\frac{\pi }{7}) + sin(x)sin(\frac{\pi }{7})=cos(x - \frac{\pi }{7})$

$cos(x - \frac{\pi }{7}) =-\frac{\sqrt{2}}{2})$

we know ,

when $cos(x) =-\frac{\sqrt{2}}{2})$ then

$x = \frac{3\pi }{4} , \frac{5\pi }{4}$

For $cos(x - \frac{\pi }{7}) =-\frac{\sqrt{2}}{2})$

$x - \frac{\pi }{7}= \frac{3\pi }{4} and x - \frac{\pi }{7}= \frac{5\pi }{4}$

Add $\frac{\pi }{7}$ on both sides

$x= \frac{3\pi }{4} + \frac{\pi }{7} and x= \frac{5\pi }{4} +\frac{\pi }{7}$

Finally we add 2npi for general solution

So options C and D are correct

6 0

3 years ago