Mathematics, 21.04.2021 18:30 kimberlyblanco14

What is the solution for x in the equation? 5
NI
3+ + 4
=
O A.
I = 4
OB.
I = -4
O C.
T =
12
7.
OD
12
1 =
-

Answers: 2

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Answers

Answer from: Quest

step-by-step explanation:

subtract 7, so the answer is x < = 3, or c.

Answer from: Quest

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,

b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,

and solve for the variable

* lets solve the problem

∵ $7^{x}=77$

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴ $ln(7^{x})=ln(77)$

- use the rule $ln(a^{n})= nln(a)$

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴ $x=\frac{ln(77)}{ln(7)}=2.23$

* the solution of the equation is 2.23 to the nearest 2 decimal places

Answer from: Quest

35=5×77=7×1when we simplify 7 by 7we will get 5 and 1, 5 divided by 1 is 5

Answer from: Quest

a) 40 more; b) old factory: 10%, new factory: 16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.

using the function, substituting 0 in for w,

$p(w)=230(1.1)^w\\\\p(0)=230(1.1)^0\\\\p(0)=230(1)=230$

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate. the equation for the old factory is $p(w)=230(1.1)^w$ ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1. this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

$\text{percent of change}=\frac{\text{amount of change}}{\text{original amount}}$

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337. this is the first week that the new factory surpasses the old.

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What is the solution for x in the equation? 5
NI
3+ + 4
=
O A.
I = 4...

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