Can someone help me in question 51 please
Answers
We know that the area of a parallelogram is the base * the height of it, so if we divide both sides by the height, then area/height=base. Therefore, we must divide the area by the height. To divide using polynomials, we first set it up similar to a regular long division problem:
______________________
2x+3 | 2x²+13x+15
Next, we take the first component of the numerator (2x² in this case) and divide it by the first component of the denominator (2x) to get x. That will form the start of our answer, and at the bottom, we will subtract our numerator by the denominator (2x+3) multiplied by the start of our answer (x). Therefore, we have
_x_____________________
2x+3 | 2x²+13x+15
-(2x²+3x)
_______________
10x+15
We then repeat the process until we finish, and whatever's left at the top is our answer. If there's something left, that's our remainder.
_x+5_____________________
2x+3 | 2x²+13x+15
-(2x²+3x)
_______________
10x+15
- (10x+15)
_________________
0
Therefore, our base has a length of x+5.
Feel free to ask further questions, and Happy Holidays!
Related Questions
I need help on this question this is really really hard.
Which shows another way to write 95?A.9 × 5B.9 + 5C.9 + 9 + 9 + 9 + 9D.9 × 9 × 9 × 9 × 9
What is the opposite of 3.78
The volumes of each figure that make the composite figure have been determined.Cone V = 24π cm3Cylinder V = 63π cm3How is the volume of the composite figure calculated?-add the 2 volumes-divide the 2 volumes-multiply the 2 volumes-subtract the 2 volumesWhat is the approximate volume of the composite figure? Round to the nearest hundredth.8.25 cubic centimeters122.52 cubic centimeters273.72 cubic centimeters4,750.09 cubic centimeters
the median of the temperatures at Meadows: ______
the interquartile range of the temperatures at Springwood:_______
the interquartile range of the temperatures at Meadows:_______
the difference of the medians as a multiple of their average interquartile range:_______
Answers
Hello,
The median of the temperatures at Springwood is 86. (median is at the middle).
the median of the temperatures at Meadows is 73. (median is at the middle).
The (IQR) interquartile range of the temperatures at Meadows is 12. (You have to subtract both of the IQR left and right 80 - 68 = 12).
The (IQR) interquartile range of the temperatures at Springwood is 14. (You have to subtract both of the IQR left and right 91 - 77 = 14).
the difference of the medians as a multiple of their average interquartile range is 79.5 is average median and 13 average IQR.
Answer: The median of the temperatures at Springwood is 86.
The median of the temperatures at Meadows is 73.
The interquartile range of the temperatures at Meadows is 12.
The interquartile range of the temperatures at Springwood is 14.
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)
Step-by-step explanation:
In a box - whisker plot , the vertical line in box represents the median value.
The left end of box denotes Lower quartile and the right end of the box Upper quartile.
By considering the given picture,
For Meadows ,
Median = 73
The median of the temperatures at Meadows is 73. (1)
Lower quartile = 68
Upper quartile = 80
Interquartile range = Upper quartile- Lower quartile
Interquartile range=80-68 = 12
The interquartile range of the temperatures at Meadows is 12. (2)
For Springwood,
Median = 86
The median of the temperatures at Springwood is 86. (3)
Lower quartile = 77
Upper quartile = 91
Interquartile range = Upper quartile- Lower quartile
= 91-77 = 14
The interquartile range of the temperatures at Springwood is 14. (4)
Difference in medians = 86-73 = 13 [From (1) and (3)]
Average interquartile range = [From (2) and (4)]
The difference of the medians as a multiple of their average interquartile range as : Difference in medians = 13= 1 x 13
i.e. Difference in medians = 1 x (Average interquartile range)
Answers
Answer:
Step-by-step explanation:
The given radical is
We want to choose from the given options, a like radical to the given expresion.
In general, the given expresion is of the form
where 'a' is a constant.
When a=1, we get the above expression.
When we substitute a=4, we get
Answers
The solution is, The area is 280.8 inches² of one triangular section.
What is area ?
Area is the measure of a region's size on a surface. The area of a plane region or plane area refers to the area of a shape or planar lamina, while surface area refers to the area of an open surface or the boundary of a three-dimensional object.
here, we have,
given that,
If the base of the panel is 31.2 inches and the height is 18 inches,
we know that,
Area of triangle = base × height ÷ 2
here, we have,
b = 31.2
h= 18
so, we get,
A = 18 × 31.2 ÷ 2
we know,
18 × 31.2 ÷ 2 = 280.8
so, A = 280.8
The area is 280.8 inches².
Hence, The solution is, The area is 280.8 inches² of one triangular section.
To learn more on Area click:
#SPJ3
18 × 31.2 ÷ 2 = 280.8
The area is 280.8 inches²
Answers
If i'm correct, the answer should be 32,918 because 5+4=9 and thus, 32,518+400=32,918
B. y = x
C. y = 5x
D. y = x + 5
Answers
We can take any two points on the line. Let's take (2, 10) and (3, 15).
Plug them into the slope formula:
m = y2-y1/x2-x1
m = 15-10/3-2
m = 5/1
m = 5
Plug it into slope-intercept form:
y = mx + b
y = 5x + b
The y-intercept here is 0, because it passes through the y-axis at 0.
So we have:
y = 5x + 0
or
y = 5x
Answers
9) obtuse 10) isosceles