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alexgriva [62]
2 years ago
6

Can you guys solve this :>

Mathematics
1 answer:
vazorg [7]2 years ago
3 0

Answer:

53˚

Step-by-step explanation:

There are 360 total degrees in a 4-sided shape (You can find the total number of degrees in a polygon by using the formula 180(n – 2) where n is the number of sides).

so we can add up all the degrees and subtract them from 360.

67 + 135 + 105 = 307

360 - 307 = 53

You might be interested in
Determine which relation is a function. Question 13 options: a) {(3, 0), (– 2, – 2), (7, – 2), (– 2, 0)} b) c) y = 15x + 2 y = 1
antiseptic1488 [7]

Answer:

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify  0-20−2  to  -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify  -2-2−2−2  to  -4−4.

3,-2,-27,-4,02y=1,5x+2d3,−2,−27,−4,02y=1,5x+2d

3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2−2d

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

5 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−4−2d

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

6 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​02y−2d

​​ ,5=x

7 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x

​1

​

​3−2d

​​ ,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

8 Simplify  \frac{3-2d}{1}

​1

​

​3−2d

​​   to  (3-2d)(3−2d).

3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

​1

​

​−2(1+d)

​​ ,5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

9 Simplify  \frac{-2(1+d)}{1}

​1

​

​−2(1+d)

​​   to  (-2(1+d))(−2(1+d)).

3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,

​1

​

​−27−2d

​​ ,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

10 Simplify  \frac{-27-2d}{1}

​1

​

​−27−2d

​​   to  (-27-2d)(−27−2d).

3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,

​1

​

​−2(2+d)

​​ ,5,

​1

​

​2(y−d)

​​ ,5=x

11 Simplify  \frac{-2(2+d)}{1}

​1

​

​−2(2+d)

​​   to  (-2(2+d))(−2(2+d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,

​1

​

​2(y−d)

​​ ,5=x

12 Simplify  \frac{2(y-d)}{1}

​1

​

​2(y−d)

​​   to  (2(y-d))(2(y−d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5=x

13 Switch sides.

x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Done

5 0
3 years ago
If (6^0)^x=1 what are the possible values of x
Monica [59]
X could equal 0 or 1
4 0
3 years ago
Given [12 -13 17 -22] [x y] = [7 -51] what is |Ay|?<br> Please help
grin007 [14]

Answer:

Step-by-step explanation:

It's not c or d. It's not much but I hope it helps.

5 0
2 years ago
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Find the intersection points of the linear and quadratic functions shown below f(x)=2x-5 g(x)= x2+2x-21
Mice21 [21]

Answer:

(-4,-13) and (4,3) the intersection points.

Step-by-step explanation:

Intersection point of two functions is a common point which satisfies both the functions.

Given functions are,

f(x)=2x-5

g(x)=x^2+2x-21

For a common point of these functions,

f(x)=g(x)

2x-5=x^2+2x-21

-5=x^2-21

0=x^2-16

x^2=16

x=-4,4

For x=-4,

f(-4)=g(-4)=2(-4)-5

                       =-13

For x=4,

f(4)=g(4)=2(4)-5

                  =3

Therefore, (-4,-13) and (4,3) the intersection points.

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