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PSYCHO15rus [73]
3 years ago
14

Jerry and Jo are making brownies for the bake sale.

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer:

Yes

12 * (12 * 18) = b will give the total number of brownies to be made

Step-by-step explanation:

Here, we want to select which of the equations will give the number of brownies to be made

The total number of brownies is b

For each pan, we have 18 rows , with 12 per row

So the total number of brownies per pan is 12 * 18

Since they are making 12, the total number of brownies b will be;

12 * (12 * 18) = b

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The sum of three consecutive odd integers is 195. Find the three integers.
Anettt [7]
Good afternoon, 

x= first odd integer 
x+2 = second odd integer
x+4= third intenger

The sum of the trhree integer is 195, so:

x + (x+2) + (x+4) = 195
3x = 195 - 4 - 2
3x=189
x= 63

Then the Answer is: 63, 65 and 67
8 0
3 years ago
Read 2 more answers
HELP !! ASAP !!
Sedaia [141]

Answer:

The roots are;

x = (2 + i)/5 or (2-i)/5

where the term i is the complex number representing the square root of -1

Step-by-step explanation:

Here, we want to use the completing the square method to solve the quadratic equation;

f(x) = -5x^2 + 4x -1

Set the function to zero

0 = -5x^2 + 4x - 1

So;

-5x^2 + 4x = 1

divide through by the coefficient of x which is -5

x^2 - 4/5x = -1/5

We now take half of the coefficient of x and square it

= -2/5^2 = 4/25

add it to both sides

x^2 - 4x/5 + 4/25= -1/5 + 4/25

(x- 2/5)^2 = -1/5 + 4/25

(x - 2/5)^2 = -1/25

Take the square root of both sides

x - 2/5 = √( -1/25

x - 2/5 = +i/5 or -i/5

x = 2/5 + i/5 or 2/5 - i/5

7 0
3 years ago
What is the limit of f (×) as x approaches - infinty
tatyana61 [14]
If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}

Hope this helps.
5 0
3 years ago
Describe a scenario where the function value exists at x=c, but the limit does not exist.
Vikki [24]

The scenario can be described using a piecewise function like:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

<h3>When the value exists but the limit does not?</h3>

Remember that the limit only exists if the limit from left and the limit from the right give the same value.

Then, we can just define a piecewise function of the form:

f(x) = 1/x                  if  x < c.

f(x) = x                    if   x = c

f(x) = 1/(x + 73)       if x > c.

Clearly, this is not a continuous function.

Notice that:

f(c) = c.\\\\ \lim_{x \to c^{-}} f(x) = 1/c\\\\ \lim_{x \to c^{+}} f(x) = 1/(c + 73)

So the limits from left and right are different, then:

\lim_{x \to c^{}} f(x)

Does not exist.

If you want to learn more about limits:

brainly.com/question/5313449

#SPJ1

6 0
2 years ago
Use mental math to figure out
never [62]
1) 94
2) 77
3) 290
4) 682
5) 45
6) 925
7) 416
3 0
4 years ago
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