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2 years ago

PLS HELPS!! THIS IS WORTH 200 PTS!! FIRST ONE ASNWERS CORRECTLY, GETS THE BRAINIEST.

Mathematics

2 answers:

Verizon [17]2 years ago

8 0

Answer:

B) −2

Explanation:

When x is positive then y is positive integer.

When x is negative then y is negative integer.

For Option A

y = 2(-3) + 5

y = -1 [x is negative so is y]

For Option B

y = 2(-2) + 5

y = 1 [x is negative but y is positive]

For Option C

y = 2(1) + 5

y = 7 [x is positive so is y]

For Option D

y = 2(2) + 5

y = 9 [x is positive so is y]

Blizzard [7]2 years ago

5 0

Answer:

-2

Step-by-step explanation:

If x is +ve then y is +ve
x is -ve then y is -ve

here x is negative y must be negative too

Let's check

y=2(-2)+5=-4+5=1

y is +ve

Option B is wrong

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Graph the inequality x>-8 using the steps

Yuki888 [10]

Hey!

Your point on the graph would be ( -8, 0 ).

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3 years ago

Is y = 3/2x +5/4 and -2x+8y=-25 a solution

nekit [7.7K]

Answer: x=-35/10

y=-4

Step-by-step explanation:

y = 3/2x +5/4

-2x+8y=-25

--------------------

replace y in second equation with: y = 3/2x +5/4

-2x+8(3/2x+5/4)= -25

-2x+12x+10= -25

10x=-35

x=-35/10

x=-3.5

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7 0

3 years ago

An 80-m tower is supported by a guy wire attached to the top of the tower. If the wire forms an

nalin [4]

The Sine or Sinθ in a right-angle triangle is the ratio of its perpendicular to its Hypotenuse. The length of the wire is 81.5 meters.

The Sine or Sinθ in a right-angle triangle is the ratio of its perpendicular to its Hypotenuse. it is given as,

Sin(θ) = Perpendicular/Hypotenuse

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The hypotenuse is the longest side of the triangle.

The length of the tower is 80 meters, while the angle of elevation is 79°. Therefore, the length of the wire will be the hypotenuse of the triangle. Therefore, the length of the wire is,

Sin(θ) = Perpendicular/Hypotenuse

Sin(79°) = 80 meter/Length of the wire

Length of the wire = 81.4973 ≈ 81.5 meters

Hence, the length of the wire is 81.5 meters.

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2 years ago

What is the answer to -29 + 30?

Agata [3.3K]

I hope this helps you

--29+30

+1

6 0

3 years ago

Read 2 more answers

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago