Ask question

Ask question

All categories

2 years ago

9

A rhombus has a perimeter of 52 inches. It’s shorter diagonal is 10 inches in length. which of the following is the length of it

s longer diagonal?

1 answer:

Korolek [52]2 years ago

4 0

Answer:

ok it's a good question... so answer to that is... I also don't know..

You might be interested in

Help me please so I can turn this in

Ber [7]

Answer:

10y-12

Step-by-step explanation:

3 0

2 years ago

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 $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

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

4 0

3 years ago

Determine the range of the following graph:

Allushta [10]

Answer:

Determine the range of the following graph:

12

11

O

10

9

8

7

6

5

NA

Step-by-step explanation:

(f+g)(x)

f(x)= x-3, g(x)= 2x+8

8 0

2 years ago

The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

4 0

3 years ago

Find the value of x.

faust18 [17]

Answer:

7 sqrt(3) =x

Step-by-step explanation:

We know that sin 60 = opposite / hypotenuse

sin 60 = x/14

Multiply both sides by 14

14 sin 60 = x

14 * (sqrt(3)/2) =x

7 sqrt(3) =x

8 0

3 years ago

Read 2 more answers