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3 years ago

Find the positive value of x in the geometric sequence 7x-2,4x+4,3x

1 answer:

5 0

Hello,

Let's assume k the ratio.

$
 \left \{ {{3x=k*(4x+4)} \atop {4x+4=k*(7x-2)}} \right. \\\\
 \left \{ {{3x-4kx=4k} \atop {4x-7kx=-2k-4}} \right. \\\\
 \left \{ {{x(3-4k)=4k} \atop {x(4-7k)=-2k-4}} \right. \\\\
 \left \{ {{x=\dfrac{4k}{3-4k} \atop {x=\dfrac{-2k-4}{4-7k} \right. \\\\

\dfrac{4k}{3-4k}=\dfrac{-2k-4}{4-7k}\\\\

36k^2-6k-12=0\\
6k^2-k-2=0\\

\boxed{k= \dfrac{2}{3} \ or\ k= -\dfrac{1}{2}}\\\\

x=\dfrac{4*\dfrac{2}{3} }{3-4*\dfrac{2}{3} }=8\\
x=\dfrac{4*\dfrac{-1}{2} }{3-4*\dfrac{-1}{2} }=-\dfrac{2}{5}\\





$

$\boxed{x= 8 \ or\ x= -\dfrac{2}{5}}\\\\$

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Simplify (-2 + 3i) + 2(5 + 6i). Enter your answer in the form a + bi.

Reptile [31]

We will have the following:

$(-2+3i)+2(5+6i)=-2+3i+10+12i$

$=(10-2)+(3i+12i)=8+15i$

So, the solution would be:

$8+15i$

3 0

1 year ago

In a small town in the Himalayas, the average snowfall each month is 6 inches. July is the month with the lowest average snowfal

OLga [1]

The function that models the above behavior is given as:

f(x) = ACos(B(x + C)) + 6 ..... .......i

where x = 1 (that is January)

D (Average Snowfall) = 6.

Hence, the amplitude (A) is given as:
A = 6-1

= 5

<h3>What is a mathematical function?</h3>

A mathematical function is a function comprising of different variables that represent different factors or events and the relationships between them.

<h3>
During what period is there more than 10 inches of snowfall?</h3><h3 />

Recall that in a year we have 12 months. Hence, P = 12

→ P = (2π) /.B → (2π)/B = 12 → B = π/6

Taking the values of A and B and plugging them in, we ahve:

F(x) = 5 cos (π/6(x + C)) + 6................ii

This means that in July, we'll have

x = 7 and f(x) = 1

Therefore,

1 = 5cos ((π/6) (7 + C) + 6 → 5 Cos ((π/6) (7 + C) → -5cos ((π/6) (7 + C)

= -1

→ ((π/6) (7 + C) = π

→ 7 + C = 6

→ C = -1

Taking equation ii, we have

F(x) = 5cos ((π6) (x -1)) + 6

F (x) > 10

Therefore

5cos ((π/6) (x-1) + 6 > 10

→ cos ((π/6) (x-1) + 6 > 4/5

→ (π/6) (x-1) > Cos⁻¹ (4/5)

(π/6) (x-1) < 0.6435 and

(π/6) (x-1) > 5.6397

→ x < 2.23 and x > 11.77

The interpretation is that upwards of ten inches of snow fall in January, February, November, and December.

Learn more about functions at;
brainly.com/question/25638609
#SPJ1

3 0

2 years ago

Find the limit , picture provided

V125BC [204]

Answer:

d. does not exist

Step-by-step explanation:

The given limits are;

$\lim_{x \to 4} f(x) =5$ , $\lim_{x \to 4} g(x) =0$ and $\lim_{x \to 4} h(x) =-2$

We want to find

$\lim_{x \to 4} \frac{f}{g}(x)= \lim_{x \to 4} \frac{f(x)}{g(x)}$

By the properties of limits, we have;

$\lim_{x \to 4} \frac{f}{g}(x)= \frac{\lim_{x \to 4} f(x)}{\lim_{x \to 4} g(x)}$

This gives us;

$\lim_{x \to 4} \frac{f}{g}(x)= \frac{5}{0}$

Division by zero is not possible. Therefore the limit does not exist.

7 0

3 years ago

56: =24:3

dmitriy555 [2]

56:7=24:3
3x3=27:3
8x2=64:4
45:9=35:7
4x5=10x2
42:6=63:9
28:4=7x1
2x2=36:9

5 0

3 years ago

Can anyone please help me with this ?

djverab [1.8K]

Answer:

The distance from point A to the top of the building is 5 $\sqrt{2}$ feet

The Height of the Skyscraper is 5 feet

Step-by-step explanation:

Given

Let Top point of building be point C

Also Let Base of the building be point B

Distance from point A to base B of the building AB= 5 feet

∴ It makes a Right angle triangle

Also ∠ACB = 45°

Also tan 45° = 1

Now tan ∠ACB = $\frac{AB}{BC}$

∴ AB= BC =5 feet

The Height of the Skyscraper is 5 feet

Now Triangle ABC is right angle triangle with right angled at B

So by Pythagoras theorem

AC= $\sqrt{AB^2+BC^2}=\sqrt{5^2+5^2} =\sqrt{50} =\sqrt{25\times2} =\sqrt{5^2\times2}=5\sqrt{2}$

The distance from point A to the top of the building is 5 $\sqrt{2}$ feet

5 0

3 years ago

Read 2 more answers