How would you write is city had no more than 5 feet of snow in one snowfall as a in inequality?

Find direction numbers for the line of intersection of the planes x + y + z = 1 and x + z = 0. (enter your answers as a comma-se

Soloha48 [4]

<1, 1, 1> × <1, 0, 1> = <1, 0, -1>

The cross product of the normals of the planes gives the direction vector of their line of intersection.

7 0

3 years ago

Given that B is the midpoint of segment AC, find the unknown values.

shusha [124]

Hi! for x, you need to do 2x-8 = x+17 and solve, because since b is the midpoint, ab and bc would be congruent.

2x-8 = x+17
1x-8 = 17
1x = 25
x = 25

in order to find ab, you plug in the x into the expression.

ab = 2x-8
ab = 2(25)-8
ab = 42

do the same for bc:

bc = x + 17
bc = 25 + 17
bc = 42

and for ac, combine the two expressions and plug in the value for x.

ac = ab + bc
ac = 2x-8 + x+17
ac = 2(25)-8 + 25+17
ac = 84

i hope this helped! have a good day/night :)

5 0

3 years ago

A rectangle has a length that is 5 meters greater than the width. If w represents the width, write an expression, in terms of w,

ahrayia [7]

Answer:

Area of rectangle is $w^2+5w$

Perimeter of Rectangle is $4w+20$ .

Step-by-step explanation:

Given:

Let the width of the rectangle be 'w'.

Also Given:

A rectangle has a length that is 5 meters greater than the width.

Length of rectangle = $w+5\ meters$

We need to write expression for Area of rectangle and Perimeter of rectangle.

Solution:

Now we know that;

Perimeter of rectangle is equal to twice the sum of the length and width.

framing in equation form we get;

Perimeter of rectangle = $2(w+5+w)=2(2w+5) =4w+20$

Also We know that;

Area of rectangle is length times width.

framing in equation form we get;

Area of rectangle= $w(w+5) = w^2+5w$

Hence Area of rectangle is $w^2+5w$ and Perimeter of Rectangle is $4w+20$ .

6 0

3 years ago

Mr. Green teaches mathematics and his class recently finished a unit on statistics. The student scores on this unit are: 40 47 5

Harrizon [31]

Answer:

Mean = 64.46, Median = 62 and Mode = Bi-modal (50 and 62)

Range of the data is 55.

Step-by-step explanation:

We are given that Mr. Green teaches mathematics and his class recently finished a unit on statistics.

<u>The student scores on this unit are:</u> 40, 47, 50, 50, 50, 54, 56, 56, 60, 60, 62, 62, 62, 63, 65, 70, 70, 72, 76, 77, 80, 85, 85, 95.

We know that Measures of Central Tendency are: Mean, Median and Mode.

Mean is calculated as;

Mean = $\frac{\sum X}{n}$

where $\sum X$ = Sum of all values in the data

n = Number of observations = 24

So, Mean = $\frac{40+ 47+ 50+ 50+ 50+ 54+ 56+ 56+ 60 +60+ 62+ 62+ 62+ 63+ 65+ 70+ 70+ 72+ 76+ 77+ 80+ 85+ 85+ 95}{24}$

= $\frac{1547}{24}$ = 64.46

So, mean of data si 64.46.

For calculating Median, we have to observe that the number of observations (n) is even or odd, i.e.;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

Now here in our data, the number of observations is even, i.e. n = 24.

So, Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(\frac{24}{2})^{th}\text{ obs.} +(\frac{24}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(12)^{th}\text{ obs.} +(13)^{th}\text{ obs.} }{2}$

= $\frac{62 + 62 }{2}$ = $\frac{124}{2}$ = 62

Hence, the median of the data is 62.

A Mode is a value that appears maximum number of times in our data.

In our data, there are two values which appear maximum number of times, i.e. 50 and 62 as these both appear maximum 3 times in the data.

This means our data is Bi-modal with 50 and 62.

The Range is calculated as the difference between the highest and lowest value in the data.

Range = Highest value - Lowest value

= 95 - 40 = 55

Hence, range of the data is 55.

5 0

3 years ago

Helpppppppppppppppppp

Wittaler [7]

8 ÷ 2 x 4 = 16

I hope I helped. If I got the right answer, Brainliest would be appreciated.

7 0

3 years ago