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4 years ago

What’s the value of x

Mathematics

2 answers:

lubasha [3.4K]4 years ago

6 0

The answer is 3radical 2

jenyasd209 [6]4 years ago

4 0

Answer:

The correct answer is last option 3√2 units

Step-by-step explanation:

From the figure we can see that two small right angled triangle.

These two triangle combined to form a large right angled triangle.

To find the value of x

Consider the small triangle with hypotenuse x and one side 3 units.

The angles of this triangle be 45°, 45° and 90°, then the sides are in the ratio 1 : 1 : √2

Therefore we can write,

3 : 3 : x = 1 : 1 : √2

x = 3√2

The correct answer is last option 3√2 units

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It is time for you to graduate from College or University! CONGRATULATIONS! The Graduation

nadezda [96]

Answer:

$150 is the independent variable and $25 is the dependent variable

Step-by-step explanation:

independent variables are do not change. dependent variables change.

4 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

A survey by the state health department found that the average person ate 208 pounds of vegetables last year and 125 5/8 pounds

Trava [24]

I think you would divide

8 0

3 years ago

12 is 8% of what number?

jeka57 [31]

We can also say this in another way ... (8% of what number is 12?)
let a be the number of which 8 % is 12 ....
8/100 x a =12 
a=12 x 100/8 
a=(6/4) x 100 
a=(3/2) X 100
a = 150 

now if u want to check your answer.. substitute the value of a .....
(8/100) x 150=12 ....

Hope it helps !!!

7 0

3 years ago

Read 2 more answers

Thank you guys for helping me :) :)

Ainat [17]

Answer:

The second one is right

Step-by-step explanation:

Hope it helpssss

4 0

3 years ago