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cricket20 [7]
3 years ago
8

What is the equation of the line whose Y intercept of three and So what is the equation of the line whose Y-intercept is 3 and s

lope is 1?
Mathematics
1 answer:
viva [34]3 years ago
3 0

Answer:

y = x + 3

Step-by-step explanation:

the equation ofa line in slope-intercept form is

y = mx + c ( m is the slope and c the y-intercept )

here m = 1 and c = 3, hence

y = x + 3 ← is the equation of the line


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Please help me with this!!
34kurt

Answer:

36

Step-by-step explanation:

Think of the unknown sized side as a half of 12 and multiply them together when your triangle is obtuse like that. and split the answer from 6 times 12.

I hope this helped you!

6 0
3 years ago
Explain the steps in finding the midpoint of the points (-1,2) and (3,-6). The state the midpoint
blondinia [14]

Answer:

(1, -2)

Step-by-step explanation:

midpoint x = (x₁ + x₂) / 2

midpoint y = (y₁ + y₂) / 2

(\frac{-1 + 3}{2} , \frac{2 + -(6)}{2} )

  = (1, -2)

8 0
3 years ago
Please help meeeee
miskamm [114]

Answer:

the graph has opposite x intercepts of b and -b. the graph has y intercepts at –b². the graph of the function symmetrical about the y-axis.

Step-by-step explanation:

3 0
3 years ago
If 2/5 divided by blank = 1/2 what is blank
cestrela7 [59]
Hey There!

To find the blank you have to multiply \frac{2}{5} * \frac{1}{2}= \frac{1}{5}
8 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}

9 +  {x}^{2}  = 25

{x}^{2}  = 16

x = 4

so our opposite side is 4. sin =Opposite/Hypotenuse so our sin a =4/5.Sin is negative in the third quadrant so

sin a =-4/5.

Now use cosine difference formula

-  \frac{3}{5}  \times  \frac{5}{13}  +   - \frac {4}{5}  \times  \frac{12}{13}

- \frac{15}{65} + (  - \frac{48}{65}  )

-  \frac{63}{65}

Hope this helps

6 0
3 years ago
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