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mr Goodwill [35]

3 years ago

13

A recipe for making 3 dozen muffins requires 1 1/2 cups of flour. How many cups of flour are required to make 5 dozen of the sam

e muffins? A) 2 cups B) 2 1/2 cups C) 3 cups D) 3 1/2 cups

1 answer:

Lunna [17]3 years ago

3 0

The answer is B. 2 1/2

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A whole number is 6 more than 2 times another number. The sum of the two numbers is less than 50. This can be written in an ineq

alukav5142 [94]

Answer:

13, 14

Step-by-step explanation:

The parameters of the numbers are;

A whole number value = 2 × Another number + 6

The sum of the two numbers is less than 50

Given that the first number is equal to more than twice the second number, we have that the first number is the larger number, while the second number is the smaller number

Where 'x' represents the second number, we get;

x + 2·x + 6 < 50

Simplifying gives;

3·x + 6 < 50

x < (50 - 6)/3 = 14. $\overline 6$

x < 14. $\overline 6$

Therefore, the numbers for which the inequality holds true are numbers less than 14. $\overline 6$ . From the given option, the numbers are 13, and 14.

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AveGali [126]

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3 years ago

Find the surface area of the regular pyramid. _____sq. cm.

snow_tiger [21]

Ashes2049, are you good at geometry?

7 0

3 years ago

Find the least number which must be added to 6203 to obtain a perfect square.

andrey2020 [161]

Answer:

79

Step-by-step explanation:

78

---------

7 | 62 03

49

----------

148 | 13 03

11 84

-----------

19

-----------

78^2 = 6084.

We observe that 78^2 < 6203.

79^2 = 6241.

We observe that 79^2 > 6203.

Hence the number to be added to 6203 is 6241 - 6203 = 38.

6203 + 38 = 3241

= 79 * 79

= 79.

Therefore 38 should be added to 6203 to obtain a perfect square.

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If $P(x) = p_nx^n + p_{n-1}x^{n-1}+\ldots+p_0$ is divided by $(x-a)$ , then $P(x) = (x-a) \cdot Q(x) + R(x)$ for some polynomials $Q,R$ . Moreover, $\deg R < 1$ (because $\deg (x-a) = 1$ ), so there exists $\alpha \in \mathbb{R}$ such that $R(x) = \alpha$ for all $x \in \mathbb{R}$ . But if we calculate $P(a)$ , it turns out that $P(a) = (a-a)\cdot Q(a) + \alpha$ , so $R(x) = \alpha = P(a)$ . $\blacksquare$

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3 years ago