Help i need good grade on this or ill never make it

What is the answer please show work 7x + 4 = 5x + 10

RSB [31]

Answer:

x=3

Step-by-step explanation:

subtract 5x from 7x

subtract 4 from 10

divide 2x and 6 by 2

7 0

3 years ago

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For each of the following sequences, • Give a formula for the nth term in the sequence, • Give a recursive definition for the se

Andrew [12]

Answer:

(a) $n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) $n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

(c) $n^{th} = n!$

$f(1) = 1$

$f(n) = f(n-1) * n$

Step-by-step explanation:

(a) This is a sequence of consecutive number

$n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

$n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

(c) This is factorial sequence. Where the next number is this number multiplied by $n^{th}$

$n^{th} = n!$

$f(1) = 1$

$f(n) = f(n-1) * n$

5 0

3 years ago

Which phrase best describes the translation from the graph y=(x-5)^2+7 to the graph of y=(x+1)^2-2

Helga [31]

Start from the parent function $f(x)=x^2$

In the first case, you are computing

$f(x-5)+7$

In the second case, you are computing

$f(x+1)-2 /tex] There are two translation going on: when you transform [tex] f(x) \to f(x+k)$ , you translate the function horizontally, $k$ units left if $k>0$ and $k$ units right if $k$ .

On the other hand, when you transform $f(x) \to f(x)+k$ , you translate the function vertically, $k$ units up if $k>0$ and $k$ units down if $k$ .

So, the first function is the "original" parabola $f(x)=x^2$ , translated $5$ units right and $7$ units up. Likewise, the second function is the "original" parabola $f(x)=x^2$ , translated $1$ units left and $2$ units down.

So, the transformation from $(x-5)^2+7$ to $(x+1)^2-2$ is: go $6$ units to the left and $2$ units down

8 0

3 years ago

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Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask

Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$

$=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3$

$=0.95019264$

8 0

3 years ago

($1.75+2times $0.25 +5times $0.05) times 24

Anastaziya [24]

(1.75 + 2 x 0.25 + 5 x 0.05) x 24= 17

8 0

3 years ago

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