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Nookie1986 [14]
3 years ago
8

Which of the following are exterior angles? Check all that apply.​

Mathematics
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

B. <4

D. <5

Step-by-step explanation:

Exterior angle is the angle found outside the triangle. In the diagram given, angle 5 and angle 4 are located outside of the triangle, therefore, the exterior angles in the diagram given are <4 and <5.

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In a game, a contestant had a starting score of one point. He tripled his score every turn for four turns. Write his score after
snow_tiger [21]
4 to the power of  1 i think thats the answer

6 0
3 years ago
Read 2 more answers
Solve this problem thanks
vovikov84 [41]

The three missing lengths are the left hypotenuse, x, the middle altitude, y, and the right hypotenuse, z.


9/y = y/16


y^2 = 9 * 16


y^2 = 144


y = 12


9^2 + 12^2 = x^2


x^2 = 225


x = 15


12^2 + 16^2 = z^2


z^2 = 400


z = 20


From left to right, the sides measure 15, 12, and 20 units.

5 0
4 years ago
Read 2 more answers
Which expression is equivalent to (5³) ‐²? (4 points)
miss Akunina [59]

Answer:

1. 2³ × 2² = (2 × 2 × 2) × (2 × 2) = 23+2 = 2⁵

2. 3⁴ × 3² = (3 × 3 × 3 × 3) × (3 × 3) = 34+2 = 3⁶

3. (-3)³ × (-3)⁴ = [(-3) × (-3) × (-3)] × [(-3) × (-3) × (-3) × (-3)]

= (-3)3+4

= (-3)⁷

4. m⁵ × m³ = (m × m × m × m × m) × (m × m × m)

= m5+3

= m⁸

Step-by-step explanation:

exponential law

3 0
3 years ago
Let u= &lt;4,3&gt;. Find the unit vector in the direction of u, and write your answer in component form.
aksik [14]
Take the vector u = <ux, uy> = <4, 3>.

Find the magnitude of u:

||u|| = sqrt[ (ux)^2 + (uy)^2]

||u|| = sqrt[ 4^2 + 3^2 ]

||u|| = sqrt[ 16 + 9 ]

||u|| = sqrt[ 25 ]

||u|| = 5

To find the unit vector in the direction of u, and also with the same sign, just divide each coordinate of u by ||u||. So the vector you are looking for is

u/||u||

u * (1/||u||)

= <4, 3> * (1/5)

= <4/5, 3/5>

and there it is.

Writing it in component form:

= (4/5) * i + (3/5) * j

I hope this helps. =)
3 0
3 years ago
1. Slove the system of equations, first by graphing, and then algebraically. (8 points) y=x-2 and y=-2x+7
Anton [14]

The solution of the equation is x=3 and y=1

Explanation:

The equations are y=x-2 and y=-2 x+7

First we shall solve the equation graphically.

The image of the graph is attached below.

This contains the solution to the system of equations.

The equations y=x-2 and y=-2 x+7 are plotted on the graph.

The intersection of these two equations are the solutions of the system of equations.

Thus, the intersection of the two equations are x=3 and y=1

Now, we shall solve the equation algebraically.

Let us solve the equation using substitution method.

Let us substitute y=x-2 in y=-2 x+7, we get,

x-2=-2x+7

Adding both sides by 2x, we have,

3x-2=7

Adding both sides by 2, we get,

3x=9

Dividing both sides by 3,

x=3

Thus, the value of x is 3.

Substituting x=3 in y=x-2, we get,

y=3-2\\y=1

Thus, the value of y is 1.

Hence, The solution of the equation is x=3 and y=1

4 0
3 years ago
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