What is the area of this triangle?

Evaluate the expression: 4(8 − 6)5 2 − 6 ÷ (−3).

mr Goodwill [35]

Answer:

86 or 32, depending on the correctness of the syntax

Step-by-step explanation:

We use PEMDAS:

Parentheses: 4(2)5 2-6/(-3)

Exponents: I'm assuming that all the syntax is correct (so no exponents).

Multiplication and Division: 40*2+6 = 80 + 6

Addition and Subtraction: 86

If you copy+pasted this expression, then the original equation would have been: (4(8 − 6)^5)/(2 − 6 ÷ (−3)). We can use PEMDAS:

Parentheses: (4(2)^5)/(2 - 6 ÷ (−3))

We solve the parentheses individually using PEMDAS:
Exponents: (4*32) <-- Numerator | Denominator -->(2 - 6 ÷ (−3))
Multiplication and Division: 128 <-- Numerator | Denominator --> 2 + 2
Addition and Subtraction: 128 < -- Numerator | Denominator --> 4

We now have the following fraction: 128/4

That simplifies to 32

7 0

3 years ago

Given the midpoint and one of the endpoints of a line segment, find the other endpoint. Midpoint: (0,3) Endpoint: (6,-3)

andrey2020 [161]

Answer:

$(-6,9)$

Step-by-step explanation:

Midpoint: (0,3)

Endpoint: (6,-3)

Use the midpoint formula:

$M=(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2})$

Since you already have the midpoint and you need an endpoint, let the unknown endpoint be (x,y). Take the midpoint formula apart:

$\frac{x_{1}+x{2}}{2}=m_{x}$

$\frac{y_{1}+y_{2}}{2} =m_{y}$

$m_{x}$ and $m_{y}$ are the coordinates of the midpoint. Enter the known values of the midpoint into the equations:

$(0_{m_{x}},3_{m_{y}})\\\\\frac{x_{1}+x_{2}}{2}=0 \\\\\frac{y_{1}+y{2}}{2}=3$

Now enter the known endpoint values:

$(6_{x_{1}},-3_{y_{1}})\\\\\frac{6+x_{2}}{2}=0\\\\\frac{-3+y_{2}}{2}=3$

Solve for x. Multiply both sides by 2:

$2*(\frac{6+x}{2})=2*(0)\\\\6+x=0$

Subtract 6 from both sides:

$6-6+x=0-6\\x=-6$

Now solve for y. Multiply both sides by 2:

$2*(\frac{-3+y}{2})=2*(3)\\\\ -3+y=6$

Add 3 to both sides:

$-3+3+y=6+3\\y=9$

Now take the values of x and y and turn into a point:

$x=-6\\y=9\\(-6,9)$

Finito.

7 0

3 years ago

3. Make a class frequency table using 2 rule of the following data. Find the Mean and make a Cumulative frequency Histogram [10]

sashaice [31]

Answer: Mean = 284

Step-by-step explanation:

Data:

125, 126, 127, 127, 137

201, 202, 207, 211, 215, 218, 224, 237, 261

301, 308, 310, 315, 330, 330, 360, 380, 393, 394

401, 401, 410, 421, 422, 424

Class Frequency Table:

$\boxed{\begin{array}{c|c|c}\underline{\text{Intervals}}&\underline{\text{\qquad Frequency\qquad}}&\underline{\text{Cumulative Frequency}}\\0-49&0&0\\50-99&0&0\\100-149&5&5\\150-199&0&5\\200-249&8&13\\250-299&1&14\\300-349&6&20\\350-399&4&24\\400-449&6&30\\450-499&0&30\end{array}}\\.\qquad \qquad \qquad \qquad \quad \ 30$