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anastassius [24]
2 years ago
14

Using the imagine below, find the measure of angle y.

Mathematics
1 answer:
koban [17]2 years ago
7 0

y+ (5x-10)= 180 º

Due to the lines are parallels

\alpha=\beta

So then

(5x-10)= (x+86)

5x - x = 86 + 10

4x = 96

x = 24

__________________

Checking

(5x-10= 5*24-10= 110º

(x+86) = 24+ 86= 110º

__________________________

y+ (5x-10)= 180 º

y+ 110º = 180 º

y= 180-110º

y= 70º

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levacccp [35]

#1

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3 0
2 years ago
In constructing a box-and-whisker plot from a data set, which values are at the ends of the box?
I am Lyosha [343]

Answer:  In constructing a box-and-whisker plot from a data set,  the minimum and maximum values are at the ends of the box.


Step-by-step explanation:

A box and whisker plot displays the five measures of data. The five measures summary is the minimum, first quartile, median, third quartile, and maximum. In a box plot, we draw a box from the first quartile to the third quartile.

Therefore, In constructing a box-and-whisker plot from a data set,  the minimum and maximum values are at the ends of the box.

Hence, A is the right option,

8 0
3 years ago
Identify the percent amount and base in this problem.150 is 25 percent of what number.
NemiM [27]

let's say that number is "x", so that is then the 100%.

if 150 is 25%, what is "x"?


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6 0
3 years ago
In January 2005, it cost $0.37 to mail a letter weighing up to 1 ounce cost $0.23. At most, what is the weight of a letter with
solniwko [45]

Answer:

3 ounces at most

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5 0
4 years ago
Match the parabolas represented by the equations with their foci.
Elenna [48]

Function 1 f(x)=- x^{2} +4x+8


First step: Finding when f(x) is minimum/maximum
The function has a negative value x^{2} hence the f(x) has a maximum value which happens when x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2. The foci of this parabola lies on x=2.

Second step: Find the value of y-coordinate by substituting x=2 into f(x) which give y=- (2)^{2} +4(2)+8=12

Third step: Find the distance of the foci from the y-coordinate
y=- x^{2} +4x+8 - Multiply all term by -1 to get a positive x^{2}
-y= x^{2} -4x-8 - then manipulate the constant of y to get a multiply of 4
4(- \frac{1}{4})y= x^{2} -4x-8
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is 12- \frac{1}{4}=11.75

Hence the coordinate of the foci is (2, 11.75)

Function 2: f(x)= 2x^{2}+16x+18

The function has a positive x^{2} so it has a minimum

First step - x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4
Second step - y=2(-4)^{2}+16(-4)+18=-14
Third step - Manipulating f(x) to leave x^{2} with constant of 1
y=2 x^{2} +16x+18 - Divide all terms by 2
\frac{1}{2}y= x^{2} +8x+9 - Manipulate the constant of y to get a multiply of 4
4( \frac{1}{8}y= x^{2} +8x+9

So the distance of focus from y-coordinate is \frac{1}{8} to the north of y=-14
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: f(x)=-2 x^{2} +5x+14

First step: the function's maximum value happens when x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25
Second step: y=-2(1.25)^{2}+5(1.25)+14=17.125
Third step: Manipulating f(x)
y=-2 x^{2} +5x+14 - Divide all terms by -2
-2y= x^{2} -2.5x-7 - Manipulate coefficient of y to get a multiply of 4
4(- \frac{1}{8})y= x^{2} -2.5x-7
So the distance of the foci from the y-coordinate is -\frac{1}{8} south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when x=8.5 and y=79.25. The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when x=-2.75 and y=-10.125. Manipulating coefficient of y, the distance of foci from y-coordinate is \frac{1}{8} to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when x=1.5 and y=9.5. The distance of the foci from the y-coordinate is \frac{1}{8} to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0
3 years ago
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