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HACTEHA [7]
3 years ago
15

Write an equation for this relationship.

Mathematics
1 answer:
disa [49]3 years ago
8 0

Answer:

y=18.5x

Step-by-step explanation:

74-55.5=18.50

92.50-74=18.50

111-92.5=18.50

(3,55.50)

y=18.5x +b

55.50=18.5(3)+b

55.50=55.50+b

b=0

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Use the graphing method to solve the equation 5(1/2)^x=30 Round to the nearest thousandth
Daniel [21]

Answer:

x = -2.585

Step-by-step explanation:

You will have to plug this into your graphing calculator.

In y=, type 5(1/2)^x in the Y1 and type 30 in Y2.

When you hit 2nd trace and hit find intersections, you will get (-2.585, 30).

The answer is x = -2.585

5 0
3 years ago
What is 8 and 2/7 + 3 and 1/3
kifflom [539]
8and2/7+3and 1/3=>(8*7+5)/7+(3*3+1)/3=>61/7+4/3=>183/21+28/21=>211/21.
5 0
3 years ago
The quotient of m and 350 equals 299
Sveta_85 [38]

Answer:

Step-by-step explanation:

m÷350 = 299

m = 299×350 = 104,650

8 0
2 years ago
In simplest radical form
azamat

√18 is answr i think

8 0
3 years ago
50 points + brainliest
xxTIMURxx [149]
Solving this problem involves repeated application of the distance formula. In order to figure out which vertices we need to connect to another vertex, we should first plot the points on the coordinate plane to get an idea of what the polygon looks like. To form the sides of this polygon (which is, in our case, a pentagon), we'll need to connect the points in the following pairs:

(-2, -2) and (3, -3)
(3, -3) and (4, -6)
(4, -6) and (1, -6)
(1, -6) and (-2, -4)
(-2, -4) and (-2, -2)

In case you forgot, the distance formula is simply an application of the Pythagorean Theorem that treats the x-distance and y-distance between two points as the "legs" of a right triangle, and the shortest distance between them as the "hypotenuse."

If a and b are the legs of a right triangle, and c is the hypotenuse, the Pythagorean Theorem can be written as:

a^2+b^2=c^2

Or, if we're just looking for the value of c:

c=\sqrt{a^2+b^2}

Since the hypotenuse in our case represents <em>distance</em>, it's more descriptive to rename that variable <em>d</em>. Also, the "legs" a and b in this problem represent the distances between the x and y components of the two points. If we take any two points (x_1,y_1) and (x_2,y_2), the distance between the x components of those points would be their difference, x_2-x_1, and the distance between the y components would be y_2-y_1. Substituting that all in, the distance formula becomes:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

All that's left to do now is substitute our specific points into the formula for each side of the polygon:

(-2, -2) and (3, -3):
d=\sqrt{(3-(-2))^2+(-3-(-2))^2}\\ d=\sqrt{(3+2)^2+(-3+2)^2}\\ d=\sqrt{5^2+(-1)^2}\\ d=\sqrt{25+1}\\ d=\sqrt{26}\\ d\approx5.1

(3, -3) and (4, -6)
d=\sqrt{(4-3)^2+(-6-(-3))^2}\\ d=\sqrt{1^2+(-6+3)^2}\\ d=\sqrt{1+(-3)^2}\\d=\sqrt{1+9}\\ d=\sqrt{10}\\ d\approx3.2

(4, -6) and (1, -6)
d= \sqrt{(1-4)^2+(-6-(-6))^2} \\ d= \sqrt{(-3)^2+0^2} \\ d= \sqrt{9} \\ d=3

(1, -6) and (2, -4)
d= \sqrt{(2-1)^2+(-4-(-6))^2}\\ d= \sqrt{1^2+(-4+6)^2}\\ d= \sqrt{1+2^2}\\ d= \sqrt{1+4} \\ d= \sqrt{5}\\ d\approx2.2

(2, -4) and (-2, -2)
d= \sqrt{(-2-2)^2+(-2-(-4))^2}\\ d= \sqrt{(-4)^2+(-2+4)^2} \\ d= \sqrt{16+2^2}\\ d= \sqrt{16+4}\\ d= \sqrt{20} \\ d\approx4.5

Rounding beforehand and adding up all of the distances gives us a perimeter of 18 units, which is remarkably close to the more precise approximation of 17.96 units. Given your options, 17.9 units would be the closest to the result we obtained here.

6 0
3 years ago
Read 2 more answers
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