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3 years ago

Find the value of x in each case:

Mathematics

2 answers:

morpeh [17]3 years ago

5 0

Answer:

x=15

Step-by-step explanation:

5x+(180-9x)+(180-8x)=180

total of a triangle angles is 180

MrRissso [65]3 years ago

3 0

Answer:

15°

Step-by-step explanation:

Step 1:

180° - 8x + 5x + 180° - 9x = 180° Sum of Triangle

Step 2:

360° - 12x = 180° Algerbra

Step 3:

- 12x = 180° Algerbra

Step 4:

12x = 180° Algerbra

Answer:

x = 15°

Hope This Helps :)

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Draw a rectangle that has a perimeter of 20 units and an area of 25 square units

marissa [1.9K]

<h2>Explanation:</h2><h2></h2>

For any rectangle the dimensions are given by:

$w:width \\ \\ l:Length$

So the perimeter (P) is given by the following equation:

$P=2(w+l)$

And the area is:

$A=w\times l$

We know that:

$P=20units \\ \\ A=25units^2$

By substituting:

$20=2(w+l) \\ \\ \therefore w+l=10 \ \ldots eq1 \\ \\ \\ 25=w\times l \ldots eq2$

Solving for $w$ from eq1:

$w=10-l$

Substituting into eq2:

$25=(10-l)\times l$

Then:

$\text{Solving for l:} \\ \\ \\ 10l-l^2=25 \\ \\ 10l-l^2-25=25-25 \\ \\ -l^2+10l-25=0 \\ \\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ a=-1,\:b=10,\:c=-25 \\ \\ \\ l_{1,\:2}=\frac{-10\pm \sqrt{10^2-4\left(-1\right)\left(-25\right)}}{2\left(-1\right)} \\ \\ l_{1,\:2}=\frac{-10\pm \sqrt{0}}{2\left(-1\right)} \\ \\ l=5$

Solving for w:

$w=10-5=5$

So this rectangle is a square and is shown below.

8 0

3 years ago

Round 5.492 to the place value of the underlined digit.

mars1129 [50]

If it has a line under the 9, then it needs to be 5.49, but if it has a line under the 4 it needs to be rounded up to 5.5 since the number behind it is above 5.

6 0

3 years ago

If <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B3%7D%7Bx%2B2%7D%20-%5Csqrt%7Bx-3%7D" id="TexFormula1" title="f(x)=\fra

ira [324]

Answer:

The domain for f(x) is all real numbers greater than or equal to 3.

Step-by-step explanation:

$f(x)= \frac{3}{x+2} - \sqrt{x-3}$

To solve the above equation, the value of x should be greater than or equal to 3, as we have the term $\sqrt{x-3}$ and if value is less than 3, the term inside the square root will be negative and we cannot find the square root of negative numbers.