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4 years ago

6

The dot plot shows the highest daily temperature recorded in a city each year for the past 15 years. What is the interquartile r

ange of the data? A/97 B/4 C/6 D/98

1 answer:

Ber [7]4 years ago

5 0

The dot plot shows the highest daily temperature recorded in a city each year for the past 15 years. What is the interquartile range of the data?

B. 4

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How to solve linear equations

Harman [31]

There are many methods to solve linear equations, such as:

1. Equating method: Let's take two simultaneous linear equations x+y=5 and x-y=1.

x+ y=5 can be written as x = 5-y.........(1)

In the same way,

x-y=1 or, x = 1+y.............(2)

Here, we equate the two equations and get the value of unknowns.

2. Substituting method: Again, take the equations x+y=5 and x-y=1.

x+y=5

or, x=5-y.........(1)

x-y=1............(2)

Substituting equation (1) in (2),

(5-y) - y=1

or, 5-2 y = 1

or, y = 2.

This is called substitution method.

3. Formula method: simultaneous linear equations can also be solved by formula.

If there are two equations: ax+by +c=0 and d x+e y+f=0 then the solution of such equation would be:

x = (bf - e c)/( a e - db), y = (a f-dc)/(db - a e).

For example,

2 x+ 3 y = 10 or, 2 x +3 y-10=0..................(1)

x+y=-44 or, x+y+44=0..................(2)

Comparing equation (1) and (2) with ax+by+c = 0 and dx+ey+f=0, a=2, b=3, c=-10, d =1, e=1, f=44.

Using formula,

x= -142 and y = 98.

4. Graphical method

5. Elimination method

6. Matrix method

8 0

4 years ago

Instructions: state what additional information is required in order to know that the triangle in the image below are congruent

Stells [14]

SSS (side-side-side) requires that all three sides of the potentially congruent triangles be congruent.

Triangle WVU & Triangle UBW are our two triangles.

The middle line, UW and WU are congruent by the reflexive property, so that is not the information we need to prove SSS.

For SSS, we need to know that sides WV and UB are congruent.

Hope this helps!

8 0

3 years ago

X^5 + 243 divided by x+3

ch4aika [34]

-3   | 1   0   0   0   0   243 $\longleftarrow$ coefficients of the polynomial you're dividing
.     |                                   $\longleftarrow$ drop down the leading coefficient
- - - - - - - - - - - - - - - - - - -
.     | 1

On the left side of the frame, we write -3 because we're dividing by $x+3=x-(-3)$ . (The algorithm is followed for division of a polynomial by a factor of $x-c$ .) Since we're dividing a degree 5 polynomial by a degree 1 polynomial, we expect to get a degree 4 polynomial.

-3   | 1   0   0   0   0   243
.     |      -3                         $\longleftarrow$ multiply -3 by 1, write in next column, add to 0
- - - - - - - - - - - - - - - - - - -
.     | 1 -3

Repeat step for the remaining columns.

-3   | 1   0   0   0   0   243
.     |      -3   9
- - - - - - - - - - - - - - - - - - -
.     | 1 -3   9

-3   | 1   0   0      0   0   243
.     |      -3   9   -27
- - - - - - - - - - - - - - - - - - -
.     | 1 -3   9   -27

-3   | 1   0   0      0 0   243
.     |      -3   9   -27   81
- - - - - - - - - - - - - - - - - - - - -
.     | 1 -3   9   -27   81

-3   | 1   0   0      0 0 243
.     |      -3   9   -27   81 -243
- - - - - - - - - - - - - - - - - - - - -
.     | 1 -3   9   -27   81       0

which translates to

$\dfrac{x^5+243}{x+3}=x^4-3x^3+9x^2-27x+81$

So the bottom row of the frame gives the coefficients of each term in the quotient by descending order. Since the last coefficient is 0, this means the remainder upon division vanishes, i.e. $x^5+243$ is exactly divisible by $x+3$ .

- - -

Another way to get the same result is to use a well-known result: for $a+b\neq0$ ,

$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)\implies\dfrac{a^5+b^5}{a+b}=a^4-a^3b+a^2b^2-ab^3+b^4$

and in this case $a=x$ and $b=3$

3 0

4 years ago

Solve the system of equations -12x-5y=40 12x-11y=80

Gwar [14]

Answer:

The answer is (-5/24, -7.5).

Step-by-step explanation:

There are two methods to solving a system of equations - substitution or elimination. In this example, we can use elimination since the coefficients of 'x' are opposite signs. To complete elmination, we simply add the equations together:

-12x - 5y = 40

<u>+ 12x - 11y = 80</u>

-16y = 120

Use inverse operations to solve for 'y' by dividing both sides by -16 to get y = -7.5.

Next, plug in the value of 'y' into one of the equations and solve for 'x':

-12x - (5)(-7.5) = 40 or -12x + 37.5 = 40, now subtract 37.5 from both sides: -12x + 37.5 - 37.5 = 40 - 37.5 or -12x = 5/2. Lastly, divide both sides by -12 to solve for x, x = -5/24.

4 0

3 years ago

Paul has three cube-shaped boxes. Each box is a different size

qwelly [4]

Answer:

See explanation

Step-by-step explanation:

Paul has three cube-shaped boxes. Each box is a different size and they are stacked from the largest to the smallest. Some information about the boxes is given below.

The combined volume of the three boxes is 1,197 cubic inches.
The area of one face of the medium box is 49 square inches.
The volume of the smallest box is 218 cubic inches less than the volume of the medium box.

1. The medium box has the area of one face of 49 square inches, then

$a^2=49\\ \\a=7\ inches$

is the side length.

The volume of the medium box is

$a^3=7^3=343\ in^3.$

2. The volume of the smallest box is 218 cubic inches less than the volume of the medium box, then the volume of the smallest box is

$343-218=125\ in^3.$

Ib is the side length, then

$b^3=125\\ \\b=5\ inches$

The area of one face is

$b^2=5^2=25\ in^2.$

3. The volume of the largest box is

$1,197-343-125=729\ in^3,$

then if c is the side length,

$c^3=729\\ \\c=9\ inches.$

4. The total height of the stack is the sum of all sides lengths:

$a+b+c=7+5+9=21\ inches$

5. Find the surface area of each box:

small $6b^2=6\cdot 5^2=6\cdot 25=150\ in^2;$
medium $6a^2=6\cdot 7^2=6\cdot 49=294\ in^2;$
large $6c^2=6\cdot 9^2=6\cdot 81=486\ in^2.$

In total, Paul needs

$150+294+486=930\ in^2$

of wrapping paper. He has 1,000 square inches, so Paul has enough paper to wrap all 3 boxes.

8 0

4 years ago