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4 years ago

The larger of two numbers is five more than twice the smaller number. The sum of the two numbers is 38. Find the numbers.

Mathematics

1 answer:

White raven [17]4 years ago

8 0

Answer:

11 and 16

Step-by-step explanation:

A good way to solve these is to work backwards.

X= larger and Y= smaller

if X is 5 more than 2Y than subtract 5 from the total

38 - 5 = 33

divide by 3

33/3 = 11

11 = Y and double that is 22 so add the five back into X and you get 16

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Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.

vovikov84 [41]

Answer:

$x=-1, S=\{x\:\in\mathbb{R}:x=-1\}$

Step-by-step explanation:

For the sake of clarity, assuming you meant:

$|x^{2}+2x+3|\le2$

1) Absolute Value or Modulus functions has one property that assures us that:

$If |a|$

2) Solving for x

$I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}$

Not defined in Real Set of Numbers

Evaluating $x^{2}+2x+3\leqslant 2$ :

$x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}$

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for $x^{2}+2x+3\geqslant -2\: \in \mathbb{C}$ . The only solution in the Real Set for the inequality $|x^{2}+2x+3|\le2$ is $S=\{-1\}$ , i.e. x=-1.

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4 years ago

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Marie is going to make a cake for her mother's birthday. It will take 38 minutes to make the cake batter, and the cake needs to

Andrej [43]

Answer:

She should start making the batter at 1:17 p.m.

Step-by-step explanation:

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3 years ago

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Aleks [24]

Answer:

15 students

Step-by-step explanation:

No. of boys students = No. of all students - No. of girls

no. of boys =50-15 = 35 boys

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and the girls that will play the match are 5 girls

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3 years ago