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2 years ago

What is the range of possible sizes for side x?

Mathematics

2 answers:

faust18 [17]2 years ago

6 0

0.5 < x < 16.5 given: Two sides of triangle: 8.0 units and 8.5 units
Measure of third side = x
According to the triangle's inequality,
Sum of any two sides > third side. (i)
Difference between the sides < third side. (ii)
If x is the third side, then
x < 8+8.5 [Using (i)]
i.e. x< 16.5
Also, x > 8.5-8 [Using (ii)]
i.e. x> 0.5
Hence, Range of possible sizes for side x would be 0.5 < x < 16.5.

Rudik [331]2 years ago

4 0

16.5 given hope thus helps

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Helpppppppppppppppppp

UNO [17]

Answer:

<em>2 solutions</em>

Step-by-step explanation:

Given the expression

2m/2m+3 - 2m/2m-3 = 1

Find the LCM of the expression at the left hand side:

2m(2m-3)-2m(2m+3)/(2m+3)(2m-3) = 1

open the bracket

4m²-6m-4m²-6m/(4m²-9) = 1

Cross multiply

4m²-6m-4m²-6m = 4m² - 9

-12m = 4m² - 9

4m² - 9+12m = 0

4m² +12m-9 = 0

<em>Since the resulting equation is a quadratic equation, it will have 2 solutions since the degree of the equation is 2</em>

8 0

2 years ago

According to the table, which ordered pair is a local minimum of the function, f(x)?

sveticcg [70]

Answer:

Option D

Step-by-step explanation:

Given is a table which gives the order pair of x and y for a function f(x)

We have to find the local minimum of the function f(x)

We have from the table the values of different f(x)

Of all we find the least value is -15 and for this x value is -2 or +2

Hence f(x) has minimum at two points

(-2,-15) and (2,-15)

Out of 4 options given we find that (2,-15) appears in IV option

Hence option D is right answer

7 0

2 years ago

Read 2 more answers

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .