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3 years ago

On square PQRS below, if Q is located at (7,0) and R is located at (5,-8) what is the length of SR

Mathematics

1 answer:

jenyasd209 [6]3 years ago

8 0

the length of SR is $2\sqrt{17}$ .

<u>Step-by-step explanation:</u>

Here we have , On square PQRS below, if Q is located at (7,0) and R is located at (5,-8) . We need to find Length of side SR . Let's find out:

We know that , In a square there are 4 sides , and length of every side is same . So side length of SR = side length of QR . Now , Let's find side length of QR .

We know that distance between two points $Q(x_1,y_1),R(x_2,y_2)$ given by :

⇒ $QR = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Two points given are $Q(7,0), R(5,-8)$ :

⇒ $QR = \sqrt{5-7)^2+(-8-0)^2}$

⇒ $QR = \sqrt{(2)^2+(-8)^2}$

⇒ $QR = \sqrt{4+64}$

⇒ $QR = \sqrt{68}$

⇒ $QR = 2\sqrt{17}$

But QR=SR , Therefore , the length of SR is $2\sqrt{17}$ .

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Answer:

(8x−3)(x+2)

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Factor 8x2+13x−6

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Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

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3 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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Answer:

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b) It is proportional. It is a straight line that passes through the origin.

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2 years ago

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Answer:

tHE FIRST OPTION

Step-by-step explanation:

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