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2 years ago

Find m Angle DEH and m Angle FEH.

Mathematics

1 answer:

horrorfan [7]2 years ago

7 0

Answer:

$13x + 10x + 21 = 90 \\ 23x + 21 = 90 \\ 23x = 69 \\ x = 3$

The measure of angle DEH

$13(3) = 39 \: degrees$

commercial angle FEH

$10(3) + 21 = \\ 30 + 21 = 51 \: degrees$

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Read 2 more answers

Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\$

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$

<u>Separate the solutions</u>

$k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2}$

The equation has complex roots

<u>Separate the real and imaginary parts</u>

We have the final answer as

$k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i$

Hope this helps you

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2 years ago

Aliyah was counting her steps. It took her 1,317 steps to walk to the store, 561 steps to get to the park, and another 259 to ma

trapecia [35]

Answer:

Aliyah will take a total of 4274 steps.

Step-by-step explanation:

Given:

Number of steps from home to store = 1317 steps

Number of steps from store to park = 561 steps

Number of steps from park to friends house = 259 steps

So now we will find the Total number of steps from home to friends house.

Total number of steps from home to friends house can be calculated by adding Number of steps from home to store and Number of steps from store to park and Number of steps from park to friends house.

framing in equation form we get;

Total number of steps from home to friends house = $1317+561+259 = 2137\ steps$

Is same route is followed to home we can say that;

Total number of steps from friends house to home = 2137 steps.

So Total number of steps she will take next day will be equal to Total number of steps from home to friends house plus Total number of steps from friends house to home.

framing in equation form we get;

Total number of steps she will take next day = 2137 +2137 = 4274 steps

Hence Aliyah will take a total of 4274 steps.

3 0

2 years ago

3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o

Fiesta28 [93]

Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true that f must be surjective? Is it true that g must be surjective? Justify your answers with either a counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not injective (since both 2, 3 7→ 4) but g ◦ f is injective. Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2 . Then g is not injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective. Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b. Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page. The part of this problem most groups had the most issue with was the second. Everyone should be comfortable with carefully proving a function is surjective by the time we get to the midterm.

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3 years ago