All categories

3 years ago

Pls tell me how to do it

Mathematics

1 answer:

Nikolay [14]3 years ago

4 0

Answer:

1) 35

2) 125

Step-by-step explanation:

angle adc = x

x+ 90 + 20 = 180 (triangke sum property)\

x = 180 - 110 = 70

angle bdc = 70/ 2 = 35 ( the median cuts the angle into 2)

2) cbd = x = 125

x + c + bdc = 180

x + 35 + 20 = 180

x = 180 - 55 = 125

I hope im right!!!

You might be interested in

F(z) = (z - 5)(2x + 7) (73 = 3) has zeros at x = -3.5, x=3/7 , x=5 What is the sign of f on the interval 3/7 < x <5?

Gwar [14]

<h3>Answer: Negative</h3>

======================================================

Explanation:

3/7 = 0.42857 approximately

Pick a number between that value and 5, not including either endpoint. Let's say we pick x = 2

Plug x = 2 into the f(x) function

f(x) = (x - 5)(2x + 7)(7x-3)

f(2) = (2 - 5)(2*2 + 7)(7*2-3)

f(2) = (2 - 5)(4 + 7)(14-3)

f(2) = (-3)(11)(11)

f(2) = -363

The actual result doesn't matter. All we're after is whether the result is positive or negative. We see the result is negative. This means f(x) is negative when 3/7 < x < 5. The f(x) curve is below the x axis on this interval.

3 0

3 years ago

For a one-tailed test with a = .05 and a sample of n = 9, the critical value for the t statistic is t = 1.860.

ahrayia [7]

I think its B. False if im wrong sorry hope this helped tho :p

6 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

2 years ago

What two whole numbers is 29−−√ between?: 3 and 4 4 and 5 5 and 6 6 and 7

jolli1 [7]

The two whole numbers are 5 and 6 ⇒ 3rd answer

Step-by-step explanation:

To prove that a square root number lies between which two consecutive integers do that

Find a square number less than the number under the root
Find a square number greater than the number under the root
Find the square root of the square numbers, they will be the two integers that the root lies between them

∵ The number is $\sqrt{29}$

- Find a square number less than 29

∵ 25 is a square number

∵ 25 is less than 29

- Find a square number greater than 29

∵ 36 is a square number

∵ 36 is greater than 29

∴ 25 < 29 < 36

- Take √ for each number

∴ $\sqrt{25}$ < $\sqrt{29}$ < $\sqrt{36}$

∵ $\sqrt{25}$ = 5

∵ $\sqrt{36}$ = 6

∴ 5 < $\sqrt{29}$ < 6

The two whole numbers are 5 and 6

Learn more:

You can learn more about the numbers in brainly.com/question/9621364

#LearnwithBrainly

8 0

4 years ago

it takes 1 hour longer to fly to st paul at 200 mi/h than it does to return at 250 mi/h. How far away is st.paul

gtnhenbr [62]

Answer:

1000 miles

Step-by-step explanation:

We will use the distance equation to find the answer to this problem.

D = RT

Where

D is distance

R is rate

T is time

At 250 miles per hour, suppose it takes t hours to go to st. paul. Thus, we can write:

D = 250t

Now, at 200 mi per hour, it takes 1 hour longer, so the time is "t + 1", so we can write:

D = 200(t+1)

We can equate both and see:

$250t=200(t+1)\\250t=200t+200$

Now, we solve for t first:

$250t=200t+200\\250t-200t=200\\50t=200\\t=\frac{200}{50}\\t=4$

We want the distance of St. Paul, so we can use any equation. lets use the 1st equation:

D = 250t

we know t = 4

D = 250 * 4 = 1000 miles

Distance to St. Paul is 1000 miles

7 0

3 years ago