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4 years ago

Without actually drawing the figure, could you form a triangle using side lengths 7, 8, and 18 units? Why or why not.

Mathematics

2 answers:

Marysya12 [62]4 years ago

7 0

Answer:

Step-by-step explanation:

ZanzabumX [31]4 years ago

6 0

Here is an example you can learn from.

#1: Is it possible to form a triangle with the given side lengths? If not, explain why not. 1. 5 cm, 7 cm, 10 cm SOLUTION:   The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5 ANSWER:   Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5

2. 3 in., 4 in., 8 in. SOLUTION:   No; . The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: 3+4*8

3. 6 m, 14 m, 10 m SOLUTION:   Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6 . The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER:   Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6

4. MULTIPLE CHOICE If the measures of two sides of a triangle are 5 yards and 9 yards, what is the least possible measure of the third side if the measure is an integer? A 4 yd B 5 yd C 6 yd D 14 yd SOLUTION:   Let x represents the length of the third side. Next, set up and solve each of the three triangle inequalities. 5 + 9 > x, 5 + x > 9, and 9 + x > 5 That is, 14 > x, x > 4, and x > –4. Notice that x > –4 is always true for any whole number measure for x. Combining the two remaining inequalities, the range of values that fit both inequalities is x > 4 and x < 14, which can be written as 4 < x < 14. So, the least possible measure of the third side could be 5 yd. The correct option is B. ANSWER:   B

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Identify the type of symmetry in<br> 1<br> Function or not a function

lisabon 2012 [21]

Is there a picture???????

5 0

3 years ago

If it took 28 minutes to run 4 miles, then at that rate, how long would it take to run 20 miles?

SSSSS [86.1K]

Answer:

140 minutes

Step-by-step explanation:

After 28 mins you've gone 4 miles, assuming this rate is constant and you dont slow down or speed up, in order to get to 20 miles you will have run 5x as much (20/4 = 5), thus 5x the amount of time (28x5).

20/4 = 5

5x28 = 140

8 0

3 years ago

Read 2 more answers

Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)

muminat

Answer:

The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Step-by-step explanation:

Given equation of line as :

5 x + 2 y = 12

or, 2 y = - 5 x + 12

or , y = $\frac{-5}{2}$ x + $\frac{12}{2}$

Or, y = $\frac{-5}{2}$ x + 6

∵ Standard equation of line is give as

y = m x + c

Where m is the slope of line and c is the y-intercept

Now, comparing given line equation with standard eq

So, The slope of the given line = m = $\frac{-5}{2}$

Again,

The other line if passing through the points (- 2 , 3 ) And is parallel to given line

So, for parallel lines condition , the slope of both lines are equal

Let The slope of other line = M

So, M = m = $\frac{-5}{2}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M x + c

Now , satisfying the points

So, 3 = $\frac{-5}{2}$ × ( - 2 ) + c

or, 3 = $\frac{10}{2}$ + c

Or, 3 = 5 + c

∴ c = 3 - 5 = - 2

c = - 2

So, The equation of line with slope $\frac{-5}{2}$ and passing through points ( -2 , 3)

y = $\frac{-5}{2}$ x - 2

or, 2 y = - 5 x - 4

I.e 5 x + 2 y + 4 = 0

<u>Similarly</u>

The other line if passing through the points (- 2 , 3 ) And is perpendicular to given line

So, for perpendicular lines condition,the products of slope of both lines = - 1

Let The slope of other line = M'

So, M' × m = - 1

Or, M' × $\frac{-5}{2}$ = - 1

Or, M' = $\frac{-1}{\frac{-5}{2}}$

Or, M' = $\frac{2}{5}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M' x + c'

Now , satisfying the points

So, 3 = $\frac{2}{5}$ × ( - 2 ) + c'

or, 3 = $\frac{- 4}{5}$ + c'

Or, 3 × 5 = - 4 + 5× c'

∴ 5 c' = 15 + 4

or, 5 c' = 19

Or, c' = $\frac{19}{5}$

So, The equation of line with slope $\frac{2}{5}$ and passing through points ( -2 , 3)

y = $\frac{2}{5}$ x + $\frac{19}{5}$

y = $\frac{2 x + 19}{5}$

Or, 5 y = 2 x + 19

Or, 2 x - 5 y + 19 = 0

Hence The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

And The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Answer

4 0

3 years ago

A road has a 10% grade, meaning increasing 1 unit of rise to every 10 units of run.

finlep [7]

Answer:

Part a) The elevation of the road is $6\°$

Part b) The rise is $0.2\ km$

Step-by-step explanation:

Part a) What is the elevation of the road to the nearest degree?

Let

y-----> the rise of the road ( vertical distance)

x ----> the run of the road (horizontal distance)

we have

y/x=1/10

we know that

The ratio y/x is equal to the tangent of the angle of the elevation of the road

Let

$\theta$ ----> angle of the elevation of the road

$tan(\theta)=y/x$

$tan(\theta)=1/10$

$\theta=arctan(1/10)=6\°$

Part b) If the road is two km long, how much does it rise?

using proportion

$1/10=y/2$

$y=2/10=0.2\ km$

3 0

3 years ago

Find the flux of the following vector fields across the given surface with the specified orientation. You may use either an expl

BlackZzzverrR [31]

The tetrahedron passes through the intercepts (5, 0, 0), (0, 2, 0), and (0, 0, 10). Parameterize the surface (call it $\Sigma$ ) by

$\vec r(u,v)=(1-v)\langle5,0,0\rangle+v\left((1-u)\langle0,2,0\rangle+u\langle0,0,10\rangle\right)$

$\vec r(u,v)=\langle5(1-v),2(1-u)v,10uv\rangle$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $\Sigma$ to be

$\vec r_v\times\vec r_u=\langle20v,50v,10v\rangle$

Then the flux of $\vec F(x,y,z)=\langle x,y,z\rangle$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\langle5(1-v),2(1-u)v,10uv\rangle\cdot\langle20v,50v,10v\rangle\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1100v\,\mathrm du\,\mathrm dv=\boxed{50}$

6 0

4 years ago