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n200080 [17]
3 years ago
7

#1

Mathematics
1 answer:
mojhsa [17]3 years ago
5 0

Answer:

6 cups to make 126 cookies

Step-by-step explanation:

dozen = 12 cookies

7 x 12= 84

126 - 84 = 42

so you will have to make 42 cookies more and that will be 2 more cups of flour

it total is 6 cups to make 126

You might be interested in
Which point is a solution to the inequality shown in this graph?
kykrilka [37]

Answer:

The only solution can be (0,-3) point.

Step-by-step explanation:

We have to judge whether the points in options are the solution to the graphed inequality or not.

The first point is (5,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The second point is (6,0) which not included in the shaded region of the graph. Hence, it can not be a solution.

The third point is (0,-5) which not included in the shaded region of the graph. Hence, it can not be a solution.

The fourth point is (0,-3). It is on the firm red line which is included in the shaded region of the graph. Hence, it is a solution.

Therefore, the only solution can be (0,-3) point. (Answer)

5 0
3 years ago
What is the value of the expression 2x² – 5x + 6 when<br> x = -2?
Svet_ta [14]

Answer:

24 is correct answer

Step-by-step explanation:

2 {x}^{2}  - 5x + 6 \\  = 2 {( - 2)}^{2}  - 5 \times ( - 2) + 6 \\  = 2 \times 4 + 10 + 6 \\  = 8 + 10 + 6  \\  = 24

hope it helped you:)

3 0
4 years ago
I don’t understand how to solve this problem??
Paraphin [41]
I think every angle below 90 degrees
7 0
3 years ago
Read 2 more answers
Which property is demonstrated?
Korolek [52]

Answer:

Multiplicative Inverse



4 0
3 years ago
If f(x) = x + 4 and g(x)=x^2-1, what is m(g o f)(x)?
Andreas93 [3]

Until now, given a function  f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given  f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and  

let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.  

 

Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of  exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find  f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

8 0
3 years ago
Read 2 more answers
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