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3 years ago

What is the area of a triangle expressed as a monomial

Mathematics

2 answers:

Jobisdone [24]3 years ago

8 0

Answer:

$28x^2$

Step-by-step explanation:

$area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2}$

zmey [24]3 years ago

6 0

Triangle area =1/2 x high x base
=1/2 X 7x X 8x= 1/2 X 56x
28x

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rachel's rectangle has either 2 pairs of dides the same length or all sides the same length? true or false

natita [175]

Answer:

false

Step-by-step explanation:

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3 years ago

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Can 5ft 9ft 14ft make a triangle?

ss7ja [257]

Answer:

Any dimensions can make a triangle, however

this does not make a right angled triangle.

Step-by-step explanation:

Determine the length of the hypotenuse of a right angle triangle with side lengths 5ft and 9ft by using the formula:

$a^2+b^2=c^2\\$

$\sqrt{a^2+b^2} = c$

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3 years ago

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What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$