How do you write 59/100 as a decimal

The equation 3sec (x) -2 = 1 solved on the interval is x = 0

From the question, we have the following parameters that can be used in our computation:

3sec (x) -2 = 1

Express sec(x) as 1/cos(x)

So, we have

3 * (1 / cos (x)) - 2 = 1

Add 2 to both sides

3 * (1 / cos (x)) = 1 + 2

Evaluate the like terms

3 (1 / cos (x)) = 3

Divide through by 3

(1 / cos (x)) = 3/3

So, we have

cos (x) =  1

Take the arc cos of both sides

x = 2nπ

On the interval [0, 2π), we have x = 0

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Answer:

B. $2492.36

Step-by-step explanation:

We are given the formula for compound interest,

[tex]A(t)=p(1+i)^x[/tex] where p is the amount of principal, i is the interest rate and x is the number of years.

In our problem, p is 2000, i = 4.5% = 4.5/100 = 0.045, and x is 5:

[tex]A(t)=2000(1+0.045)^x\\\\A(5)=2000(1.045)^5\approx 2492.36[/tex]

Maria will have [tex]\$\ 2492.36[/tex] in her account after [tex]5[/tex] years.

Compound interest is the interest which is received on the principal and the previous interest.

Amount [tex]=P(1+\frac{R}{100})^t[/tex]

We have,

Invested principal [tex]=\$2000[/tex]

Compound interest Rate [tex]=4.5\%[/tex]

Time of investment [tex]=5[/tex] years

So,

Using the above mentioned formula;

Amount [tex]=P(1+\frac{R}{100})^t[/tex]

              [tex]=2000(1+\frac{4.5}{100})^5[/tex]

              [tex]=2000(1+\frac{9}{200})^5[/tex]

              [tex]=2000\ *\ (\frac{209}{200})^5[/tex]

Amount [tex]=\$\ 2492.36[/tex]

So, Amount in Maria's account is [tex]\$\ 2492.36[/tex].

Hence, we can say that Maria will have [tex]\$\ 2492.36[/tex] in her account after [tex]5[/tex] years.

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After rotating angle Q, which initially measures 30 degrees, by 15 degrees, the new angle Q' measures 45 degrees.

When we talk about rotating an angle, we are referring to moving the angle around its vertex through a specified measure. If we have angle Q which measures 30 degrees and we rotate it by 15 degrees, we are effectively adding 15 degrees to the original angle. Thus, the measure of angle Q' will be 30 degrees plus 15 degrees which equals 45 degrees. Rotations do not change the original angle but instead give us a new angle based on the original position plus the rotation.

Final answer:

A computer that uses 4W of power during 'vampire' mode over weekends throughout a year produces approximately 4.992 kg of CO2 annually. This is calculated by multiplying the total hours in vampire mode by the power consumption and then by the CO2 emission factor for electricity.

Explanation:

To calculate the amount of carbon dioxide produced from the vampire energy used by a computer that is not unplugged, we first have to establish the power consumption while in 'vampire' mode. The power consumed by the computer's LED when it is turned off but not unplugged is stated to be about 4 watts (W).

A year has 52 weeks, and each weekend amounts to 60 hours of the computer being in 'vampire' mode. To find the total number of hours in a year this mode occurs, we multiply 60 hours by 52, which gives us 3,120 hours of energy consumption annually when the computer is in 'standby' mode.

Next, to convert the total power consumed in watts to kilowatt-hours (kWh), we divide the power in watts by 1,000 (since 1 kW = 1,000 W), and then multiply by the total hours per year.

The carbon dioxide emissions can be estimated using an average emission factor for electricity production. For example, if we assume that the emission factor is about 0.4 kilograms of CO2 per kWh (which varies based on local energy sources), we can calculate the total emissions for the computer's vampire energy consumption.

Calculations: 4 W ÷ 1,000 = 0.004 kW
0.004 kW × 3,120 hours/year = 12.48 kWh/year
12.48 kWh/year × 0.4 kg CO2/kWh = 4.992 kg CO2/year

Therefore, the computer produces approximately 4.992 kg of CO2 annually from vampire energy when not unplugged over weekends.

Here are a bunch of CORRECT answers. Your answer is in the first pic. I got number 3 wrong, but it still showed the correct answer.

The value of the expression ¹²P₃ and ¹⁰C₄ will be 1,320 and 210, respectively.

A permutation is an act of correctly arranging things or pieces. Combinations are a method of taking stuff or pieces from an assortment of items or sets in which the order of the items is irrelevant.

The value of the expression ¹²P₃ is given as,

¹²P₃ = 12! / (12 - 3)!

¹²P₃ = 12 x 11 x 10 x 9! / 9!

¹²P₃ = 1,320

The value of the expression ¹⁰C₄ is given as,

¹⁰C₄ = 10! / [(10 - 4)! x 4!]

¹⁰C₄ = 10 x 9 x 8 x 7 x 6! / (6! x 4 x 3 x 2 x 1)

¹⁰C₄ = 10 x 3 x 7

¹⁰C₄ = 210

The value of the expression ¹²P₃ and ¹⁰C₄ will be 1,320 and 210, respectively.

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Final answer:

Sarah has 1 1/5 cups of nuts and must add 2/5 cup of dried fruit to make a total of 1 3/5 cups of trail mix.

Explanation:

Imagine Sarah is making a trail mix for a hiking trip. She already has 1 1/5 cups of nuts and wants to add more dried fruit to reach a total of 1 3/5 cups of trail mix. To find out how much dried fruit she should add, we can set up the equation:

1 1/5 + x = 1 3/5

Converting mixed numbers to improper fractions gives us:

6/5 + x = 8/5

Subtracting 6/5 from both sides, we find that x (the amount of dried fruit to be added) is:

x = 8/5 - 6/5

x = 2/5

Therefore, Sarah needs to add 2/5 cup of dried fruit to the nuts to make 1 3/5 cups of trail mix.

Final answer:

The expression 2d(3) is simplified by multiplying 2 by 3, yielding 6, and then by d, resulting in 6d.

Explanation:

To simplify the expression 2d(3), you simply need to multiply the variable d by the number 2 and then multiply the result by the number inside the parentheses, which is 3. This is straightforward multiplication, similar to the distributive property used in algebra to simplify expressions.

The simplified form of the expression is found by multiplying 2 by 3, giving us 6, and then multiplying that result by d. Hence, 2d(3) simplifies to 6d.

This figure is made up of a rectangle BDFE and parallelogram ABCD.

The area of this figure= Area of BDFE+Area of ABCD

Area of parallelogram=base*height and Are of rectangle= length*Width

So, Area of ABCD= CD*BH

And, Area of BDFE=BD*DF

area of this figure= BD*DF+CD*BH

From the figure CD=4 and BH=3

To find BD, Let us apply Pythagorean theorem, [tex] hypotenuse^{2} =adjacent^{2}+ opposite^{2} [/tex], to ΔBHD

[tex] BD^{2} =HD^{2}+ BH^{2} [/tex]

HD=2 and BH=3

[tex] BD^{2} =2^{2}+ 3^{2} [/tex]

[tex] BD^{2} =4+ 9=13 [/tex]

[tex] BD=\sqrt{13} [/tex]

To find DF, Let us apply Pythagorean theorem, [tex] hypotenuse^{2} =adjacent^{2}+ opposite^{2} [/tex], to ΔBHD

[tex] DF^{2} =DG^{2}+ GF^{2} [/tex]

DG=6 and GF=9

[tex] DF^{2} =6^{2}+ 9^{2} [/tex]

[tex] DF^{2} =36+ 81=117 [/tex]

[tex] BD=\sqrt{117}=3\sqrt{13} [/tex]

[tex] BD=\sqrt{13} and DF=3\sqrt{13} [/tex]

area of this figure= BD*DF+CD*BH

area of this figure= [tex] \sqrt{13} [/tex]*3[tex] \sqrt{13} [/tex]+4*3

area of this figure=13*3+12

area of this figure=39+12=51

Area of this figure=51 square units