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Ilia_Sergeevich [38]

3 years ago

15

Please help me asap

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

The second option

Step-by-step explanation:

Given

f(x) = - 6 $(1.05)^{x}$

The y- intercept is where the graph crosses the y- axis, where the x- coordinate is zero.

Let x = 0 and solve for y

f(0) = y = - 6 $(1.05)^{0}$ → note $a^{0}$ = 1, thus

y = - 6 × 1 = - 6 ← y- intercept of f(x)

From the table for g(x)

When x = 0 , y = g(x) = - 3 ← y- intercept of g(x)

Thus the y- intercept of f(x) is equal to 2 times the y- intercept of g(x)

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A ladder is 61 inches tall. How tall is it in feet and inches

professor190 [17]

Answer:

5'1

Step-by-step explanation:

Every 12 inches is a foot. 12 x 5 is 60. Plus 1. Is 61.

That mean's it's 5 feet 1 inch.

3 0

3 years ago

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Please help me find the slope between the two points m=

abruzzese [7]

Answer: m= 5/2

explanation:
5- - 5 = 10
-4 - - 8 = 4
10/4= 5/2

8 0

3 years ago

the sum of the ages of a man and his wife is at least 104 years if the man is x years old and the wife is x - 4 years old find t

nirvana33 [79]

Answer:

x + ×-4 > or =104

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am I right please ?

8 0

3 years ago

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Let $M$ be the least common multiple of $1, 2, \ldots, 20$. How many positive divisors does $M$ have

Veronika [31]

Consider the prime factorization of 20!.

$20! = 20 \times 19 \times 18 \times \cdots \times 3 \times 2 \times 1$

The LCM of 1, 2, ..., 20 must contain all the primes less than 20 in its factorization, so

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times m$

where $m$ is some integer not divisible by any of these primes.

Compare the factorizations of the remaining divisors of 20!, and check off any whose factorizations are already contained in the product of primes above.

$4 = 2^2$ - missing a factor of 2

$6 = 2\times3$ - ✓

$8 = 2^3$ - missing a factor of 2²

$9 = 3^2$ - missing a factor of 3

$10 = 2\times5$ - ✓

$12 = 2^2\times3$ - missing a factor of 2

$14 = 2\times7$ - ✓

$15 = 3\times5$ - ✓

$16 = 2^4$ - missing a factor of 2³

$18 = 2\times3^2$ - missing a factor of 3

$20 = 2^2\times5$ - missing a factor of 2

From the divisors marked "missing", we add the necessary missing factors to the factorization of $M$ , so that

$M = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 2^3 \times 3$

Then the LCM of 1, 2, 3, …, 20 is

$M = 2^4 \times 3^2 \times 5 \times7 \times 11 \times 13 \times17 \times 19$

$\implies \boxed{M = 232,792,560}$

3 0

2 years ago

Given the geometric sequence tn with t1=96 and r=−1/4, find t4.

leva [86]

Answer:

3/2

Step-by-step explanation:

Given

$t_1 = 96$

$r = \frac{1}{4}$

Required

Find t4

The nth term of a geometric sequence is calculated as:

$t_n = t_1 * r^{n-1$

So:

$t_4 = t_1 * r^{4-1$

$t_4 = t_1 * r^3$

This gives:

$t_4 = 96 * \frac{1}{4}^3$

$t_4 = 96 * \frac{1}{64}$

Express as a single fraction

$t_4 = \frac{96}{64}$

$t_4 = \frac{3}{2}$

<em>Hence, the 4th term is 3/2</em>

3 0

3 years ago