True Statements Plz help me

What's an expression that represents 5 less than the product of 7 and a number?

Butoxors [25]

Answer:

7x-5

Step-by-step explanation:

7 0

3 years ago

Jenna bought a golf club that was marked down 20% from an original price of $75. If she paid 5% sales tax, what was the total co

Oksana_A [137]

$63. She pays $60 for the club and $3 in taxes.

5 0

3 years ago

Read 2 more answers

A games shop sells marbles

Taya2010 [7]

To start, the question is asking for surface area of a cylinder, meaning you will have to use this formula to find it: A=2πrh+2πr^2.

Since you already know that the diameter of the plastic cylinder is 4 marbles long, and each marble is 2 cm in diameter, you would multiply 4 and 2 to get that your diameter is 8 cm. However, since the surface area of a cylinder formula uses radius, you would simply divide 8 by 2 to get that the radius of the cylinder is 4 cm.

Now, since you already know that the height of the cylinder is equal to 10 marbles, multiply the amount of marbles by 2 to get that the height of the cylinder is 20 cm.

Now that you know the height and the diameter of the cylinder, plug the values into the formula: A=2π(4)(20)+2π(4)^2.
Assuming that pi is 3.14, simplify the equation as according to PEMDAS and you get that the area of plastic needed to make one cylinder is 602.88 cm squared

5 0

3 years ago

Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AC = 20 and DC = 5, what is the length of BC? (Note: the fi

11111nata11111 [884]

Answer:

21.2 or 15 sqrt of 2 (PETHAGOREOM theorem)

Step-by-step explanation:

a^2+b^2=c^2

20^2+5^2=C^2

400+25=C^2

450=C^2

sqrt 450=

15 sqrt of 2 or 21.2

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