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joja [24]
1 year ago
9

The line y=−2x+7 together with the coordinate axes forms a triangle. Determine the area of the triangle

Mathematics
1 answer:
Flura [38]1 year ago
3 0

Solution:

Given that;

The line y=−2x+7 together with the coordinate axes forms a triangle.

Using a graphing tool, the graph of the line is shown below

From the graph above,

The height, h, of the triangle formed is 7 units

The base, b, of the triangle formed is 3.5 units

To find the area, A, of a triangle, the formula is

\begin{gathered} A=\frac{1}{2}bh \\ A\text{ is the area} \\ b\text{ is the base} \\ h\text{ is the height} \end{gathered}

Substitute the values of b and h into the formula above

\begin{gathered} A=\frac{1}{2}bh \\ A=\frac{1}{2}\times7\times3.5=12.25\text{ units}^2 \\ A=12.25\text{ units}^2 \end{gathered}

Hence, area, A, of the triangle is 12.25 square units

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S=(n-2)180
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647.2 + 12.88 how to answer that step by step
IgorC [24]

Answer:

660.08

Step-by-step explanation:

1. first you will need to line up the decimals and make sure the places are correct

 647.2

+   12.88

2. Then you just add them

4 0
3 years ago
On his first day of school, Kareem found the high temperature in degrees Fahrenheit to be 76.1°. He plans to use the function to
wolverine [178]
The function for conversions from Fahrenheit to Celsius is:

Celsius = (Fahrenheit - 32) x 5/9 

C(76.1) = (76.1 - 32) x 5/9
             = (44.1) x 5/9
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Therefore, 76.1 degrees Fahrenheit = 24.5 degrees Celsius. 
5 0
3 years ago
Read 2 more answers
Help please<br><br>What is 2/3÷1/2​
Luba_88 [7]

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3 0
2 years ago
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
3 years ago
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