Select all terms that apply to the shape

Consider the equation below

s344n2d4d5 [400]

Answer: B

Step-by-step explanation:

Let's solve the given equation. Then we can see which other equation has the same answer.

-5n+31=-14n-5

36=-9n

n=-4

We need to fina another equation that gives n=4 as the answer.

A. Incorrect

3-6n+3n=-3+4-4n

3-3n=1-4n

n=-2

B. Correct

$\frac{11}{4} n+2=-3+\frac{3}{2}n$

$\frac{5}{4}n=-5$

$n=-4$

C. Incorrect

1=0.75n+3.25

-2.25=0.75n

n=-3

D. Incorrect

-1-22n=-20n-9

8=2n

n=4

7 0

3 years ago

Given log4(3) =0.7925 and log4(5) =1.1610, use log properties to find log4(300).

Ede4ka [16]

Answer:

4.1145

Step-by-step explanation:

You can use the following formula to solve this kind of problems,

$log_{a}(b) + log_{a}(c) = log_{a}(bc)$

So, let's start of with log4_(300). Since the question have already gave log4_(3) =0.7925 and log4_(5) =1.1610. It means that you have to split the 300 into the simplest form where almost all of the numbers are 3 and 5.

$log_{4}(300) = log_{4}(4 \times 5 \times 3 \times 5) \\ = log_{4}(4) + log_{4}(5) + log_{4}(3) + log_{4}(5) \\ = 1 + 1.1610 + 0.7925 + 1.1610 \\ = 4.1145$

5 0

3 years ago

Solve 3x to the <br> second power <br> -2x-8=0 using an algebraic method

MatroZZZ [7]

Answer:

$x=2\textrm{ or } x=-\frac{4}{3}$

Step-by-step explanation:

Given:

The equation is given as:

$3x^{2}-2x-8=0\\3x^{2}-6x+4x-8=0\\3x(x-2)+4(x-2)=0\\(x-2)(3x+4)=0\\\\x-2=0\textrm{ or }3x+4=0\\x-2+2=0+2 \textrm{ or }3x+4-4=0-4\\x=2\textrm{ or }3x=-4\\x=2\textrm{ or}\\x=-\frac{4}{3}$

Therefore, the possible values of $x$ are 2 and $-\frac{4}{3}$

7 0

3 years ago

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

3 years ago

"For which value of c does Limit of f (x) as x approaches c not exist?"

Damm [24]

The limit does not exist at the jump discontinuity at <em>x</em> = -2.

From the left, the green-ish curve approaches 4; from the right, the orange curve approaches 6. These one-sided limits are not equal, so the two-sided limit does not exist.

7 0

3 years ago