These angles, 110º and 7xº, are examples of supplementary angles.
Supplementary angles are angles that, when added together, produce a sum of 180º.
This means 110º + 7xº = 180º.
With this equation, 110º + 7xº = 180º, we can isolate the variable and solve for x to answer this question.
110º + 7xº = 180º
The first step to solve for x is to subtract 110º from both sides of the equation.
110º + 7xº - 110º = 7xº
180º - 110º = 70º
7xº = 70º
Since we want 1xº instead of 7xº, we need to divide both sides by 7.
7xº / 7 = xº
70º / 7 = 10º
xº = 10º
Now we have our solution!
The value of x is 10º.
Let's check this to be sure.
7xº, when added to 110º, must equal 180º since 7xº and 110º are supplementary angles.
7xº + 110º = 180º
We found that x is 10º.
7(10)º + 110º = 180º
7 • 10 = 70
70º + 110º = 180º
180º = 180º
This works! That means that x = 10º is in fact the correct answer.
Answer:
x = 10º
Hope this helps!
Answer:
The first one is coefficient (A)
the second one is C
and the last one is D
if i am right or helpful plz mark as brainliest
The distance<u> between </u>Jays hair and the caffe shop is 2.5 km.
Since the mall is located 2.4 kilometers<u> north</u> of jays house and the caffe shop is located 0.7 kilometers <u>east </u>of the mall, both directions are perpendicular. So, the distance between Jays hair and the caffe shop forms the hypotenuse side of this right angled triangle.
So, using Pythagoras' theorem, we find the hypotenuse side of this triangle, L.
Pythagoras' theorem states that the sum of squares of the sides of a right angled triangle equal the square of the hypotenuse side.
So, L = √[(2.4 km)² + (0.7 km)²]
L = √[(5.76 km² + 0.49 km²]
L = √(6.25 km²)
L = 2.5 km
So, the distance<u> between </u>Jays hair and the caffe shop is 2.5 km.
Learn more about Pythagoras' theorem here:
brainly.com/question/8171420
Can you please provide the diagram.
<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u> </u>√15 units
Step-by-step explanation:
Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .
As we know ,
Distance(D) = √(x^1-x^2) +(y^1-y^2)
Now,
D= √(x^1-x^2) +(y^1-y^2)
= √(6-1) +(1+9)
= √5+10
= √15 units
: Therefore the distance between (6,1) and (1,-9) is √15 units.
