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4 years ago

Find the distance between the points (-4,-5) and (3,-5)

Mathematics

1 answer:

nika2105 [10]4 years ago

8 0

Answer:

Step-by-step explanation:

If you look at this graph, you will understand!

If you don't please ask me whenever!

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What is 3/7 in words

Marrrta [24]

Three sevenths !!!!!!

4 0

3 years ago

Passing through(-1,8) and parallel to the line whose equation is 8x-9y-2=0

notka56 [123]

<h3>Answer:</h3>

y = 8/9x + 80/9

<h3>Step-by-step explanation:</h3>

<u>We are given;</u>

Equation of a line as 8x -9y -2 = 0
Coordinates (-1,8)

We are required to determine the equation of a line parallel to the given line and passing through a point (-1,8)

<h3>Step 1: Determine the Gradient of the given line </h3>

When an equation is written in the form y = mx + c, m is the gradient.
Therefore; we could write the equation 8x -9y -2 = 0 in the form of y= mx + c

9y = 8x -2

y = 8/9x - 2/9

Therefore, the slope, m = 8/9

<h3>Step 2: Determine the equation of the line</h3>

We need to know that parallel lines have the same gradient
Therefore, the slope of the line in question is 8/9
It passes through a point (-1, 8)

We can therefore, determine the equation;

Taking another point, (x,y)

$\frac{y-8}{x+1} = \frac{8}{9}$

9(y-8) = 8(x+1)

9y - 72 = 8x + 8

9y = 8x + 8 +72

9y = 8x + 80

y = 8/9x + 80/9

Therefore, the equation of the line is y = 8/9x + 80/9

4 0

4 years ago

A figure is broken into a rectangle and a triangle. The triangle has a base of 2 and two-thirds feet and height of 3 feet. The r

ohaa [14]

Answer:

$b=2\frac23\:\sf ft$

Area of triangle = 4 ft²

Area of rectangle = $6\frac23\: \sf ft^2$

Area of irregular figure = $10\frac23\: \sf ft^2$

Step-by-step explanation:

$\begin{aligned}\sf Base\:of\:triangle\:(b) & = 5-2\frac13\\\\& = \dfrac{15}{3}-\dfrac73\\\\ & = \dfrac83\\\\ & = 2\frac23\:\sf ft \end{aligned}$

$\begin{aligned}\sf Area\:of\:a\:triangle & =\dfrac12 \sf \times base \times height\\\\& = \dfrac12 \times b \times 3\\\\ & = \dfrac12 \times \dfrac83 \times \dfrac31\\\\ & = \dfrac{24}{6}\\\\ & = 4\: \sf ft^2\end{aligned}$

$\begin{aligned}\sf Area\:of\:rectangle& =\sf length \times width\\\\& = 5 \times 1\frac13\\\\ & = \dfrac51\times \dfrac43\\\\& = \dfrac{20}{3}\\\\ & = 6\frac23\: \sf ft^2\end{aligned}$

$\begin{aligned} \sf Area\:of\:irregular\:figure & = \sf area\:of\:triangle+area\:of\:rectangle\\\\ & = 4 + 6\frac23\\\\ & = 10\frac23\: \sf ft^2\end{aligned}$