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3 years ago

14 divide by 38.36. please help asp

Mathematics

2 answers:

son4ous [18]3 years ago

7 0

Just use a calcut0.36

nika2105 [10]3 years ago

3 0

The answer is 0.364963503649635 but if you shorten it it’s 0.36 you just need a calculator!

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a group of employees bought a get-well gift for a coworker. the gift cost $46.60. if there are 22 employees in the group how muc

Solnce55 [7]

2.11$$ because 46.60 divided by 22 is 2.11

8 0

3 years ago

Quadrilateral MNPQ is the same shape but a different size than quadrilateral M'N'P'Q'

I am Lyosha [343]

If the shape is the same, but the size is different, <em>dilation is involved</em>. There may be other transformations that are also involved (such as rotation, translation, or reflection), so MNPQ may or may not be a straight dilation of M'N'P'Q'.

_____

Translation and dilation can sometimes be combined to a dilation about a point that is not the origin. Reflection might be accommodated by using a negative dilation factor. However, rotation cannot be accomplished using dilation alone.

6 0

3 years ago

Find the equation of a plane that is perpendicular to the vector −4i⃗ −4j⃗ −k⃗ and passing through the point (−2,−5,5)

vredina [299]

hello ....

the equation of a plane that is ; ax+by+cz +d =0

the vector perpendicular to this plane is : V(a,b,c)

in this exercice ; a = -4 b= -4 c = -1

then: the equation of a plane that is ; -4x-4y-z +d =0

but the plane passing through the point (−2,−5,5) :

-4(-2)-4(-5)-(5) +d =0

23+d =0

d =-23

the equation of a plane is : -4x-4y-z-23 =0

5 0

4 years ago

Let N be the smallest positive integer whose sum of its digits is 2021. What is the sum of the digits of N + 2021?

kondor19780726 [428]

Answer:

$10$ .

Step-by-step explanation:

See below for a proof of why all but the first digit of this $N$ must be " $9$ ".

Taking that lemma as a fact, assume that there are $x$ digits in $N$ after the first digit, $\text{A}$ :

$N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}$ , where $x$ is a positive integer.

Sum of these digits:

$\text{A} + 9\, x= 2021$ .

Since $\text{A}$ is a digit, it must be an integer between $0$ and $9$ . The only possible value that would ensure $\text{A} + 9\, x= 2021$ is $\text{A} = 5$ and $x = 224$ .

Therefore:

$N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}$ .

$N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}$ .

$N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}$ .

Hence, the sum of the digits of $(N + 2021)$ would be $6 + 2 + 2 = 10$ .

Lemma: all digits of this $N$ other than the first digit must be " $9$ ".

Proof:

The question assumes that $N\!$ is the smallest positive integer whose sum of digits is $2021$ . Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of $N$ is not " $9$ ".

For example: $N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}$ , where $\text{A}$ , $\text{P}$ , $\text{B}$ , and $\text{C}$ are digits. (It is easy to show that $N$ contains at least $5$ digits.) Assume that $\text{B} \!$ is one of the non-leading non-" $9$ " digits.

Either of the following must be true:

$\text{P}$ , the digit in front of $\text{B}$ is a " $0$ ", or
$\text{P}$ , the digit in front of $\text{B}$ is not a " $0$ ".

If $\text{P}$ , the digit in front of $\text{B}$ , is a " $0$ ", then let $N^{\prime}$ be $N$ with that " $0\!$ " digit deleted: $N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}$ .

The digits of $N^{\prime}$ would still add up to $2021$ :

$\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}$ .

However, with one fewer digit, $N^{\prime} < N$ . This observation would contradict the assumption that $N\!$ is the smallest positive integer whose digits add up to $2021\!$ .

On the other hand, if $\text{P}$ , the digit in front of $\text{B}$ , is not " $0$ ", then $(\text{P} - 1)$ would still be a digit.

Since $\text{B}$ is not the digit $9$ , $(\text{B} + 1)$ would also be a digit.

let $N^{\prime}$ be $N$ with digit $\text{P}$ replaced with $(\text{P} - 1)$ , and $\text{B}$ replaced with $(\text{B} + 1)$ : $N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}$ .

The digits of $N^{\prime}$ would still add up to $2021$ :

$\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}$ .

However, with a smaller digit in place of $\text{P}$ , $N^{\prime} < N$ . This observation would also contradict the assumption that $N\!$ is the smallest positive integer whose digits add up to $2021\!$ .

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this $N$ other than the first digit must be " $9$ ".

Therefore, $N$ would be in the form: $N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}$ , where $\text{A}$ , the leading digit, could also be $9$ .

6 0

3 years ago

Write the equation of the line that contains the given point and has the given slope. (15, 31), slope is 0.5

Anna11 [10]

Y=.5x+23.5 is your answer :D

7 0

4 years ago