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2 years ago

Evaluate 3ab for a=2 and b=3

Mathematics

2 answers:

photoshop1234 [79]2 years ago

8 0

Answer:

<u><em>Answer is below</em></u>

Step-by-step explanation:

Evaluate for a=2,b=3

$(3)(2)(3)\\(3)(2)(3)\\=18$

vampirchik [111]2 years ago

4 0

Answer:

Step-by-step explanation:

For this problem, we will simply plug in the values of a and b into the respective variables in the expression 3ab to "evaluate" the expression.

a = 2; b = 3

3ab

Note, that when variables like a and b are smashed with a constant, the use of multiplication is in play.

3ab = 3 * a * b

With this in mind, let's plug in the values of a and b into the expression.

3ab

= 3 (2) (3)

= 3 (6)

= 18

Hence, 3ab evaluated when a=2 and b=3 is 18.

Cheers.

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Math question #1 please show steps

-Dominant- [34]

Answer:

Step-by-step explanation:

An approximation of an integral is given by:

$\displaystyle \int_a^bf(x)\, dx\approx \sum_{k=1}^nf(x_k)\Delta x\text{ where } \Delta x=\frac{b-a}{n}$

First, find Δx. Our a = 2 and b = 8:

$\displaystyle \Delta x=\frac{8-2}{n}=\frac{6}{n}$

The left endpoint is modeled with:

$x_k=a+\Delta x(k-1)$

And the right endpoint is modeled with:

$x_k=a+\Delta xk$

Since we are using a Left Riemann Sum, we will use the first equation.

Our function is:

$f(x)=\cos(x^2)$

Therefore:

$f(x_k)=\cos((a+\Delta x(k-1))^2)$

By substitution:

$\displaystyle f(x_k)=\cos((2+\frac{6}{n}(k-1))^2)$

Putting it all together:

$\displaystyle \int_2^8\cos(x^2)\, dx\approx \sum_{k=1}^{n}\Big(\cos((2+\frac{6}{n}(k-1))^2)\Big)\frac{6}{n}$

Thus, our answer is C.

*Note: Not sure why they placed the exponent outside the cosine. Perhaps it was a typo. But C will most likely be the correct answer regardless.

5 0

2 years ago

Tavon has a gift card for $165 that loses $4 for each 30-day period it is not used. He has another gift card for $145 that loses

sweet [91]

Answer:

Equation 1 = Equation 2

165 - 4x = 145 - 3.50x

Step-by-step explanation:

Let x represent the number of 30-day periods.

Tavon has a gift card for $165 that loses $4 for each 30-day period it is not used.

Therefore the equation =

$165 -$4 × x

= 165 - 4x.......... Equation 1

He has another gift card for $145 that loses $3.50 for each 30-day period it is not used.

$145 -$3.50 × x

= 145 - 3.50x........... Equation 2

Hence, an equation for the number of 30-day periods until the value of the gift cards will be equal is obtained by equating Equation 1 and Equation 2 together

So, we have

Equation 1 = Equation 2

165 - 4x = 145 - 3.50x

We simplify further:

165 - 145 = -3.50 + 4.0x

20 = 0.5x

x = 20/0.5

x = 40

Therefore, number of each 30-day periods until the value of the gift cards will be equal is 40

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