Identify the property

Beth is cooking pizzas and cuts each one into 8 pieces. If she has 14 people coming to her house for dinner and she thinks that

Gennadij [26K]

She would make 7 pizzas

4 0

3 years ago

Read 2 more answers

John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21 About how far does

Agata [3.3K]

Answer:

John ski down the mountain is 1285.37 feet.

Step-by-step explanation:

Given : John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21.

To find : About how far does John ski down the mountain ?

Solution :

We draw a rough image of the question for easier understanding.

Refer the attached figure below.

According to question,

Let AB be the height of mountain i.e. AB=1200 feet

The angle of depression is 21 i.e. $\theta=21^\circ$

We have to find how far does John ski down the mountain i.e. AC = ?

Using trigonometric,

$\cos\theta = \frac{AB}{AC}$

$\cos(21)= \frac{1200}{AC}$

$AC=\frac{1200}{\cos(21)}$

$AC=1285.37$

Therefore, John ski down the mountain is 1285.37 feet.

7 0

4 years ago

Help I'm already way behind?

kirill [66]

Ez so person 1 and person 2 are in a bike race. person 1 starts at the 10 ft line and travels 4 ft per second. person 2 starts at 15 ft line but travels 3 ft per second.

3 0

3 years ago

Find values of a, b, and c (if possible) such that the system of linear equations has a unique solution, no solution, and infini

Cerrena [4.2K]

Answer:

IMPOSSIBLE

Step-by-step explanation:

First we set the equation system:

$x+y+0z=0\\0x+4y+z=0\\4ax+by+cz=0$

Now we set the matrix in order to have a solution for the system:

$\left[\begin{array}{ccc}1&1&0\\0&4&1\\4a&b&c\end{array}\right]$

Now we are going to apply Gauss-Jordan to find the solution of the system in terms of a, b and c:

$-4aR_{1}+R_{3}\rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&(-4a+b)&c\end{array}\right]$

Next step:

$(4a-b)R_{2}+4R_{3} \rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&0&(4a-b+c)\end{array}\right]$

Next step:

$(4a-b+c)R_{2}-R_{3} \rightarrow R_{2}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]$

Next step:

$4(4a-b+c)R_{1}-R_{2} \rightarrow R_{1}\\\\{\left[\begin{array}{ccc}4(4a-b+c)&0&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]$

With this solution, we have a new equation system:

$4(4a-b+c)=0\\4(4a-b+c)=0\\4a-b+c=0$

This system can be solved by Cramer's rule, by finding the matrix determinant:

$\left[\begin{array}{ccc}16&-4&4\\16&-4&4\\4&-1&1\end{array}\right]$

$\Delta s= (-64-64-64)-(-64-64-64)=0$

As the determinant is zero, we can say that the second system is imposible to solve.

4 0

3 years ago

Is y = -4x + 15 parallel to 8x + 2y = 7

anygoal [31]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago