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Iteru [2.4K]
3 years ago
8

2. Which equation represents a line that is perpendicular to the line represented by y = 2/3 + 1

Mathematics
1 answer:
Orlov [11]3 years ago
3 0
We need to see the graphs btw
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HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!
In-s [12.5K]

Answer:

I believe the answer could be c

4 0
2 years ago
7 cos x+1=6 secx solve for x
jonny [76]
7cos(x) + 1 = 6sec(x)
7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
[7cos(x) - 6][cos(x) + 1] = 0
cos(x) = 6/7 , x = arccos(6/7) and
cos(x) = -1, x = 180
6 0
2 years ago
Triangle OPQ is similar to triangle RST. Find the measure of side RS. Round your answer to the nearest tenth if necessary. OPQR
Andrews [41]

Answer:

RS  =  75.3

Step-by-step explanation:

You need to set up the proportion of corresponding sides.

\frac{PQ}{ST}  =  \frac{PO}{RS}

\frac{7}{31}  = \frac{17}{RS}

7(RS) = 31(17)

         =  527

RS = 527/7 = 75.2857...

     =  75.3

3 0
3 years ago
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
Degger [83]

Answer:

So do you already have the answer? Also 1/20 looks right.

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Find the slope please
zavuch27 [327]

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the slope of this graph; we are provided with two points:

\star~\mathrm{(6,0)}

\star~\mathrm{(0,-5)}

\triangle~\fbox{\bf{KEY:}}

  • What we need in order to find the slope are two points and the slope formula.

Here's the formula:

\star~\boxed{\mathrm{\cfrac{y2-y1}{x2-x1}}}

Where

y2 and y1 are y-coordinates

x2 and x1 are x-coordinates

Substitute the values:

\star~\mathrm{\cfrac{-5-0}{0-6}}}

Simplify!

\star~\mathrm{\cfrac{-5}{-6}}}

Simplify more!

\star~\mathrm{\cfrac{5}{6}}

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

4 0
1 year ago
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