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3 years ago

Please Help....................................

Mathematics

1 answer:

vazorg [7]3 years ago

5 0

The answer are:
(-4,5)

(2,2)

(-1, -3)

(3,4)

(2,-1)

(-5,6)

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On Sunday, a local hamburger shop sold a combined total of 378 hamburgers and cheeseburgers. The number of cheeseburgers was sol

ipn [44]

Answer:

The number of hamburgers sold was 126.

Step-by-step explanation:

Let h = number of hamburgers sold

let c = cheeseburgers sold

c+h = 378

c = 2h

Substitute c =2h into c+h =278

c+h =378

2h+h = 378

Combine like terms

3h = 378

Divide by 3

3h/3 =378/3

h =126

The number of hamburgers sold was 126.

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3 years ago

What is the solution to the equation 1/3(x-2)=1/5(x+4)+2 ?

Airida [17]

7.3 is the answer i believe

3 0

3 years ago

If one bottle of soda costs .99, how much do three bottles cost? how much do they cost with 5% tax added on?

bija089 [108]

3 sodas cost $2.97 +.15 tax total $3.12

8 0

3 years ago

G(x) = 2x – 4<br> h(x)=x²-3<br> Find (g.h)(5)

Mariulka [41]

Answer:

Step-by-step explanation:

This is asking you to plug in the value of h(5) into g(x). First solve for h(5)

x^2-3 = h(x)

5^2-3= 22

h(5) = 22

h(5) is 22. So now, plug that into g(x).

g(h(5)) = 2(22)-4

44-4=40

(g.h)(5) is 40.

4 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago