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vovikov84 [41]
2 years ago
9

Please help I will give you brainiest!!!

Mathematics
1 answer:
rusak2 [61]2 years ago
8 0

9514 1404 393

Answer:

  x-intercept: -14/9

  y-intercept: 7

Step-by-step explanation:

This is one of the easiest forms for finding intercepts. To find the x-intercept, set y=0 and divide both sides by the coefficient of x.

  -9x = 14 . . . . . set y=0

  x = -14/9 . . . . divide by -9

__

To find the y-intercept, set x=0 and divide both sides by the coefficient of y.

  2y = 14 . . . . set x=0

  y = 7 . . . . . . divide by 2

The x-intercept is -14/9; the y-intercept is 7.

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Please help i dont understand this at all
tamaranim1 [39]
I have no idea but I googled and I got 0.6 I hope this helps!
5 0
3 years ago
The following are scores for students in Ms. Kennedy's math class.What is the range of the scores?
Eddi Din [679]

Answer:

The answer will be D. 181

Step-by-step explanation:

603 - 422 = 181

<em>Please mark me as the brainliest.</em>

3 0
1 year ago
Find two consecutive even integers whors sum is 86. Which of the following could be used to solve the problem?
Yuliya22 [10]

Answer:

42 and 44

Step-by-step explanation:

The even numbers are numbers which can be written as 2n where n is any integer number.

Therefore, two consecutive numbers would be "2n" and "2n + 2"

We then have that

2n + (2n+2) = 86

4n + 2 = 86

4n = 84

n = 21

And the even number 2n is 2(21) = 42.

(We can verify this answer by doing 42 + 44 = 86)

5 0
2 years ago
The area of a rectangular field is 6052 m².
Aleonysh [2.5K]

Answer:

<h2>length = 89 m</h2>

Step-by-step explanation:

Area of a rectangle = l × w

where

l is the length

w is the width

From the question

Area = 6052 m²

width = 68 m

To find the length substitute the values into the above formula

6052 = 68l

Divide both sides by 68

We have the final answer as

<h3>length = 89 m</h3>

Hope this helps you

5 0
3 years ago
Read 2 more answers
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).





5 0
3 years ago
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