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3 years ago

Which of the expressions are equivalent to the one below? Check all that apply.

Mathematics

1 answer:

sdas [7]3 years ago

8 0

Step-by-step explanation:

We need to find an expression that is equivalent to 4×(8+3)

We know that,

4×(8+3) = 4(11) = 44

Option (1).

(4 • 8) + 3 = 32+3 = 35. It is incorrect

Option (2).

(8 + 3) • 4 = (11) 4 = 44. It is correct.

Option (3).

4 • (3 + 8) = 4(11) = 44. It is correct.

Option (4).

4 • 8 + 4 • 3 = 32+12 = 44. It is correct.

Hence, the correct options are (2), (3) and (4).

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Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2

kakasveta [241]

Answer:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

Step-by-step explanation:

So we have the integral:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx$

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

$y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx$

Therefore, the integral is:

$=2\int \sec(y)\tan^5(y)dy$

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

$=2\int \sec(y)\tan^4(y)\tan(y)dy$

Now, we can use a variation of the trigonometric identity:

$\tan^2(y)+1=\sec^2(y)$

So:

$\tan^2(y)=\sec^2(y)-1$

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

$=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy$

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

$u=\sec(y)\\du=\sec(y)\tan(y)dy$

Substitute:

$2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du$

Expand the binomial:

$=2\int u^4-2u^2+1du$

Spilt the integral:

$=2(\int u^4 du+\int-2u^2du+\int +1du)$

Factor out the constant multiple:

$=2(\int u^4du-2\int u^2du+\int(1)du)$

Reverse Power Rule:

$=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))$

Simplify:

$=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)$

Distribute the 2:

$=\frac{2u^5}{5}-\frac{4u^3}{3}+2u$

Substitute back secant for u:

$=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)$

And substitute back 1/2x for y. Therefore:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)$

And, finally, C:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

And we're done!

7 0

3 years ago

Read 2 more answers

Nick drew a scale drawing of a hotel. The scale he used was 7 inches = 3 feet. What is the scale factor of the drawing?

jeyben [28]

Answer:

7 : 36

Step-by-step explanation:

Nick drew a drawing of a hotel on a certain scale. The scale he used was 7 inches = 3 feet.

Now, 3 feet is equivalent to (3 × 12) = 36 inches.

Therefore, the scale of the drawing that he had used will be 7 inches to 36 inches.

Hence, the scale of the drawing will be 7 : 36 (Answer)

8 0

3 years ago

A pizzeria sells pizzas in individual slices of in pies of 8 slices. On a certain day, the pizzeria sold a total of 364 slides,

galben [10]

84 slices were sold individually.
364 - 84

The rest were sold as a pie
(364 - 84) ÷ 8

Answer D (But it should either be in a bracket or write as a fraction with the numerator as 364 - 84)

4 0

3 years ago

The temperature in degrees Celsius is 273.15 less than the temperature in kelvin. Andrew is conducting a science experiment wher

Stella [2.4K]

Answer:

The whole number temperatures, in kelvin, at which Andrew can conduct his experiment are 287, 288, 289,290,291.

Step-by-step explanation:

The temperature in degrees Celsius is 273.13 less than the temperature in kelvin; mathematically this means:

$k=c+273.15$

Where $k$ is the temperature in kelvin and $c$ is the temperature in Celsius.

From this relationship we convert Andrew's temperature range—

13.5°c to 18.5°c—to kelvin:

$k_1=13.5^oc+273.15=286.65$

$k_2=18.5^oc+273.15=291.65$

Thus Andrew can conduct his experiment between 286.65 and 291.65 kelvin, and the whole number temperatures between these extremes are 287, 288, 289,290,291.

Thus the whole number temperatures, in kelvin, at which Andrew can conduct his experiment are 287, 288, 289,290,291.

3 0

3 years ago

A vending machine sells chips at $0.55 and candy at $0.75. last month, the vending machine yielded $170.00 with the sale of 260

diamong [38]

We know that X + Y = 260 and that 0.55X + 0.75Y = 170 So X = 260 - Y now plug it in the other equation and solve <span>or try to multiply first equation with -0.55 and then aggregate with another one. hope this helps :)</span>

7 0

3 years ago