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3 years ago

Holly missed 7 out of 185 school days this year. Approximately what percent of school days was holly in attendance?

1 answer:

7 0

It 96%
subtract 7 from 185 that'll equal 178
use the 178 as the numerator for the days she attended
and 185 as the denominator as the days in total.
178/18= x/100 then solve for x and youll get 96 over 100 or 96%

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Solve for the value of x.x^2 + 4x + 4 = 0

Georgia [21]

Answer:

-2

Step-by-step explanation:

Lets factor

(x+2)(x+2)=0

x=-2

5 0

3 years ago

Fourteen less children’s meals were served than adult meals at a barbecue. Children plate were 1.50 each and adult plates were 2

Annette [7]

Answer:

134 children and 120 adult plates were served

Step-by-step explanation:

let the children mean be x

Let the adult meal be y

If fourteen less children’s meals were served than adult meals at a barbecue, then;

y = x - 14 .... 1

IF Children plate were 1.50 each and adult plates were 2.00 each with a total of 441 in amount then;

1.5x + 2y = 441 .... 2

Substitute 1 into 2;

1.5x + 2(x-14) = 441

1.5x+2x-28 = 441

3.5x = 441+28

3.5x = 469

x = 469/3.5

x = 134

Recall that y = x - 14

y = 134-14

y = 120

Hence 134 children and 120 adult plates were served

7 0

3 years ago

Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$