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Evgesh-ka [11]
4 years ago
12

Write an equation for the nth term of an arithmetic sequence then find a 25.

Mathematics
2 answers:
Damm [24]4 years ago
5 0
Look up math way and type it in its free and it will help alot.
I am Lyosha [343]4 years ago
3 0
19.

an<span> = 3n + 2

...........................

</span>
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Write the equation in slope-intercept form y−6=−2/5(x−10)
stellarik [79]

Answer: y = -2/5x + 10

Step-by-step explanation:

y − 6 = − 2/5 (x − 10)

do distributive property

y - 6 = -2/5x + 4

add six to both sides

y = -2/5x + 10

boom the answer :v

5 0
2 years ago
Determine the value of a so that A = 2i+aj+k and B = 4i-2j-2k are perpendicular​
Digiron [165]
In order to make them perpendicular, their dot product must result in zero. so:

4 \times 2 - 2a - 2 \times 1 = 0

2a = 8 - 2 = 6

a = 3
7 0
3 years ago
Solve for the variable in the following equation: 1 / 6c + 1 / 3 = - 2
tankabanditka [31]

Answer: C. C=-14

Step-by-step explanation:

Given

(1/6)c+1/3=-2

Multiply both sides by the LCM of 1/6 and 1/3(Least Common Multiple)

6[(1/6)c+1/3]=(-2)(6)

c+2=-12

Subtract 2 on both sides

c+2-2=-12-2

c=-14

Hope this helps!! :)

Please let me know if you have any questions

4 0
3 years ago
Read 2 more answers
What is 5 divide 630
Vanyuwa [196]
5 divided by 630 is about 0.007936
7 0
3 years ago
How many different right triangles are there with a hypotenuse of lenght 5 cm
Sveta_85 [38]

Answer:

Draw a horizontal line across a page of paper, somewhere in the middle. Mark a point A on the line, toward the left margin.

Spread open a compass to make good sized circle, but so that if the point of the compass is on the point you drew, the pencil fits on the page, both below the top edge and to the left of the right edge of the page. Draw the arc, from roughly the 12 O’Clock position over and down to the intersection of the line segment, at the 3 O’Clock position.

Call the opening of your compass, the radius of the arc you just drew, “5 units”.

Pick any point on the arc between the 12 and 3 positions, B. Drop a line down from that point B, perpendicular to the original horizontal line. Label the point that it intersects the horizontal line, C.

ABC is a right triangle with hypotenuse 5. Do this again with a point a little closer to the 3 O’Clock position. It’s another right triangle with hypotenuse 5.

Indeed, as you get closer to the right, along the arc, the height of the triangle declines, but the width of the triangle increases. The hypotenuse remains 5 in all cases. We picked points on the circle. We could have picked points on the horizontal line first, and raised perpendicular lines until they intersected the circle. Each point forms a distinct right triangle.

There are as many possible right triangles as there are possible points in a line segment.

Step-by-step explanation:

6 0
3 years ago
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