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3 years ago

A point has coordinates (0, 3). Where is it located in the coordinate plane?

Mathematics

1 answer:

igomit [66]3 years ago

7 0

Answer:

Option B y-axis

Step-by-step explanation:

we have the point (0,3)

The x-coordinate of the point is 0

The y-coordinate of the point is 3

Remember that

All points with x-coordinate equal to zero belong to the y-axis and all points with y-coordinate equal to zero belong to the x-axis

therefore

The point is located in the y-axis

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Please help! Which expression is equal to..?

Alenkasestr [34]

Answer:

Step-by-step explanation:

Before we can add the 2 fractions we require them to have a common denominator.

Multiply the numerator/ denominator of the first fraction by x

Multiply the numerator/ denominator of the second fraction by (x + 3)

= $\frac{3x(x)}{x(x+3)}$ + $\frac{(x-2)(x+3)}{x(x+3)}$

Now simplify the numerator, leaving the denominator

= $\frac{3x^2+x^2+x-6}{x(x+3)}$

= $\frac{4x^2+x-6}{x(x+3)}$ → A

5 0

3 years ago

I need help please I don’t understand

defon

9514 1404 393

Answer:

1. ∠EDF = 104°

2. arc FG = 201°

3. ∠T = 60°

Step-by-step explanation:

There are a couple of angle relationships that are applicable to these problems.

the angle where chords meet is half the sum of the measures of the intercepted arcs
the angle where secants meet is half the difference of the measures of the intercepted arcs

The first of these applies to the first two problems.

1. ∠EDF = 1/2(arc EF + arc UG)

∠EDF = 1/2(147° +61°) = 1/2(208°)

∠EDF = 104°

2. ∠FHG = 1/2(arc FG + arc ES)

128° = 1/2(arc FG +55°) . . . substitute given information

256° = arc FG +55° . . . . . . multiply by 2

201° = arc FG . . . . . . . . . subtract 55°

3. For the purpose of this problem, a tangent is a special case of a secant in which both intersection points with the circle are the same point. The relation for secants still applies.

∠T = 1/2(arc FS -arc US)

∠T = 1/2(170° -50°) = 1/2(120°)

∠T = 60°

3 0

3 years ago

Given cosα = −3/5, 180 < α < 270, and sinβ = 12/13, 90 < β < 180

torisob [31]

I got

$- \frac{6 3}{65}$

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

${12}^{2} + {y}^{2} = {13}^{2}$

$144 + {y}^{2} = 169$

$25 = {y}^{2}$

$y = 5$

so our adjacent side is 5.

Cosine is adjacent/hypotenuse so our cos b=5/13.

Using the interval it gave us, Angle a must be in the third quadrant. Since cos is adjacent/hypotenuse and we are given cos a=-3/5. We are going to set up an equation using pythagorean theorem,

$( - 3) {}^{2} + {x}^{2} = {5}^{2}$