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3 years ago

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

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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

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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

2 years ago