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3 years ago

Solve 2 cube root 8x+9=5

1 answer:

7 0

2 times the cube root of 8x+9=5
divide both sides by 2
cube root of 8x+9=2.5
cube everybody (put to 3rd power)
8x+9=2.5^3
8x+9=15.625
minus 9 both sides
8x=6.625
divie both sides by 8
x=0.828125

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Help please

ValentinkaMS [17]

Because none of the points on the number line are above 0, we know that the points will be negative.
This eliminates the second and third choices, as they contain positive numbers.
The first point, A, is a number less than -0.5 or -1/2
The first choice matches A with -5/16, which is less than -0.5 or -1/2
Therefore, the fourth choice must be the correct answer.

5 0

3 years ago

You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the

Otrada [13]

Answer:

Please check the explanation.

Step-by-step explanation:

a) How much will you have at the middle of the first year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

substituting the values

$A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}$

$A=300\cdot \frac{2.06}{2}$

$A=\frac{618}{2}$

$A=309$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

so substituting the values

$A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}$

$A=300\cdot \frac{2.06^2}{2^2}$

$A=318.27$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

7 0

3 years ago

A 5 kW solor energy system can collect 9000 kWh of energy per year. In Arizona, electricity cost eleven cents per kilowatt hour.

Licemer1 [7]

Answer:

19338 kWh

Step-by-step explanation:

In Arizona, electricity cost eleven cents per kilowatt-hour.

The average home uses 66 million BTU per year i.e. 66000000 BTU per year.

Now, it is given in the question that 1 BTU is equivalent to 0.000293 kWh.

So, the average home uses (66000000 × 0.000293) = 19338 kWh per year.

Therefore, the number of kilowatt-hours used per year will be 19338. (Answer)

6 0

3 years ago

A square garden is 12 m long. A hedge wall 1 meter wide surrounds

sammy [17]

I think it only surrounds 1 meter because it’s a meter long

4 0

3 years ago

e) Given the following: Let X = {1, 2, 3, 4) and a relation R on X as R= {(1,2), (2,3), (3,4)}. Find the reflexive and transitiv

Naddika [18.5K]

Answer:

The answer is $\{(1,1),(2,2),(3,3),(4,4),(1,2)(2,3)(3,4),(1,3),(1,4)\}$

Step-by-step explanation:

Remember that a reflexive relation $R\subset \mathcal{P}(X)$ , where $\mathcal{P}(X)$ is the power set of $X$ , is one which conteins the ordered pairs of the form $(a,a)$ , for $a\in X$ .

So, As the reflexive and transitive closure of $R$ (that we will denote by $\overline{R}$ ) is in particular reflexive, we must add to $R$ the elements $\{(1,1) , (2,2),(3,3),(4,4) \}$

A transitive relation $R$ is one in which if the pair $(a,b)$ and the pair $(b,c)$ are in there, then the pair $(a,c)$ must be there too.

So, to complete the relation $R$ to be reflexive and transitive we must add the pair $(1,3)$ (because $(1,2),(2,3)$ are in $R$ ), the pair $(2,4)$ , and the pair $(1,4)$ because we added the pair $(2,4)$ .

Therfore we have that $\overline{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2)(2,3)(3,4),(1,3),(1,4)\}$ .

6 0

3 years ago