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ser-zykov [4K]
2 years ago
10

Please help me solve for x

Mathematics
2 answers:
skad [1K]2 years ago
7 0
<h3>Answer :- </h3>

  • X = 7

<h3>Solution :- </h3>

<u>If a ray stands on a line, then the sum of two adjacent angles so formed is 180 . </u>

  • 7x + 1 + 130 = 180
  • 7x + 131 = 180
  • 7x = 180 - 131
  • 7x = 49
  • x = 49/ 7
  • x = 7

<em>Hope </em><em>it </em><em>helps </em><em>~</em>

kkurt [141]2 years ago
3 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

The shown pair of angles are Co interior angle pair, therefore the sum of those angles is 180°

\qquad \sf  \dashrightarrow \: 130 \degree + 7x + 1 \degree = 180 \degree

\qquad \sf  \dashrightarrow \: 131 \degree + 7x = 180 \degree

\qquad \sf  \dashrightarrow \:  7x= 180 \degree - 131 \degree

\qquad \sf  \dashrightarrow \:  7x= 49 \degree

\qquad \sf  \dashrightarrow \:  x=49 \degree \div 7

\qquad \sf  \dashrightarrow \:  x=7 \degree

The required value of x is 7°

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Step-by-step explanation:

The <u>actual values</u> are shown on the given graph as <u>blue points</u>.

The <u>line of regression</u> is shown on the given graph as the <u>red line</u>.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43).  It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

<u>In 2000, the actual rainfall was </u><u>7</u><u> centimeters below what the model predicts</u>.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm,  shown by point (2003, 40).  It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

<u>In 2003, the actual rainfall was </u><u>4</u><u> centimeters above what the model predicts.</u>

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2 years ago
What is the inverse to "if i listen to this song, then it will get stuck in my head"
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3 0
2 years ago
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?.
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The train will pass a bridge in 40 seconds.

<h3>What is the distance?</h3>

Distance is defined as the product of speed and time.

Given that A train 360 m long is running at a speed of 45 km/hr.

Speed of train = 45 km/hr.

And it passes a bridge 140 m long.

Length of train = 360 meters

Since 1 km = 1000 meters.

and 1 hour = 3600 seconds

we have to determine the length of the train

So 1(km/hr.) = (1000/3600) = (26/72) meter/seconds

45 km/hr = 45 ×5/18 = 25/2 meter per second

So the speed of the train = 25/2 meters per second

Total distance to be covered = (360+140)m = 500m.

So time will it pass a bridge = distance/speed

Substitute the values of distance and speed in the above formula,

⇒ 500/(25/2)

⇒ 500×2/25

⇒ 40 seconds

Hence, the train will pass a bridge in 40 seconds.

Learn more about the distance here:

brainly.com/question/13269893

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8 0
2 years ago
Unit Activity: Geometric Transformations and Congruence
Llana [10]
Task 1: criteria for congruent triangles

a. 
(SSA) is not a valid mean for establishing triangle congruence. In this case we know  <span>the measure of two adjacent sides and the angle opposite to one of them. Since we don't know anything about the measure of the third side, the second side of the triangle can intercept the third side in more than one way, so the third side can has more than one length; therefore, the triangles may or may not be congruent. In our example (picture 1) we have a triangle with tow congruent adjacent sides: AC is congruent to DF and CB is congruent to FE, and a congruent adjacent angle: </span>∠CAB is congruent to <span>∠FDE, yet triangles ABC and DEF are not congruent. 

b. </span><span>(AAA) is not a valid mean for establishing triangle congruence. In this case we know the measures of the three interior sides of the triangles. Since the measure of the angles don't affect the lengths of the sides, we can have tow triangles with 3 congruent angles and three different sides. In our example (picture 2) the three angles of triangle ABC and triangle DEF are congruent, yet the length of their sides are different.
</span>
c. <span>(SAA) is a valid means for establishing triangle congruence. In this case we know </span>the measure of a side, an adjacent angle, and the angle opposite to the side; in other words we have the measures of two angles and the measure of the non-included side, which is the AAS postulate. Remember that the AAS postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent. Since SAA = AAS, we can conclude that SAA is a valid mean for establishing triangle congruence.

Task 2: geometric constructions

a. Step 1. Take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw another circle with radius AB but this time with center at B.
Step 3. Mark the two points, C and D, of intersection of both circles. 
Step 4. Use the points C and D to mark a point E in the circle with center at A.
Step 5. Join the points C, D, and E to create the equilateral triangle CDE inscribed in the circle with center at A (picture 3).

b. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. The point B is the first vertex of the inscribed square.
Step 3. Draw a diameter from point B to point C.
Step 4. Set a radius form point B to point D passing trough A, and draw a circle.
Step 5. Use the same radius form point C to point E using the same measure of the radius BD from the previous step. 
Step 6. Draw a line FG trough were the two circles cross passing trough point A.
Step 7. Join the points B, F, C, and G, to create the inscribed square BFCG (picture 4).

c. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw the diameter of the circle BC.
Step 3. Use radius AB to create another circle with center at C.
Step 4. Use radius AB to create another circle with center at B.
Step 5. Mark the points D, E, F, and G where two circles cross.
Step 6. Join the points C, D, E, B, F, and G to create the inscribed regular hexagon (picture 5).





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