Solve the equation 5x + 7y = 35 for y.

Mathematics

2 answers:

vredina [299]3 years ago

6 0

Answer:

B. $y= 5 - \frac{5}{7} x$

Step-by-step explanation:

okay, so you are trying to get y alone on one side. The first thing you want to do is try to get 7y alone on one side. You do that my subtracting 5x on both sides. You get 7y = 5x +35

Then you divide 7 on both sides and get y = -5/7x + 5

erastovalidia [21]3 years ago

4 0

Answer:

D. y = -5/7x + 5

Step-by-step explanation:

5x + 7y = 35

-5x -5x

7y = 35 - 5x

/7 /7 /7

y = 35/7 - 5/7 x

y = 5 - 5/7x

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2 years ago

I need a real answer to this problem and if you can't answer then please don't answer this. Just give me the answer for the blan

Harlamova29_29 [7]

To divide the triangles into these regions, you should construct the <u>perpendicular bisector</u> of each segment.
These perpendicular bisectors intersect and divide each triangle into three regions.
The points in each region are those closest to the vertex in that <u>region</u>.

<h3>What is a triangle?</h3>

A triangle can be defined as a two-dimensional geometric shape that comprises three (3) sides, three (3) vertices and three (3) angles only.

<h3>What is a line segment?</h3>

A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.

<h3>What is a perpendicular bisector?</h3>

A perpendicular bisector can be defined as a type of line that bisects (divides) a line segment exactly into two (2) halves and forms an angle of 90 degrees at the point of intersection.

In this scenario, we can reasonably infer that to divide the triangles into these regions, you should construct the <u>perpendicular bisector</u> of each segment. These perpendicular bisectors intersect and divide each triangle into three regions. The points in each region are those closest to the vertex in that <u>region</u>.

Read more on perpendicular bisectors here: brainly.com/question/27948960

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