50 POINTS I NEED THIS IM CONFUSED

Can some one help me

svp [43]

Answer:

Domain: All real numbers, except for x=4

Range: All real numbers

Step-by-step explanation:

There is no limit to how far either of the graphs go, so it will be infinite.

4 0

3 years ago

PLEASE HELLPPPPPPPPPPPPP!!!!!!!!

Slav-nsk [51]

The first answer is 25x + 45 = c
The second one is y=3x
Hopes this helps

6 0

2 years ago

Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

7 0

3 years ago

What is the surface area of a right rectangular prism with the length of 10 inches width of 8 inches and a height of 1.5 inchess

Contact [7]

Answer: Option 3

Step-by-step explanation: Option (3) is correct.The surface area of a rectangular prism of given dimension is 170 square inches.Step-by-step explanation:Given : a rectangular prism whose base is 5 inches by 6 inches and whose height is 5 inches.We have to find the surface area of a rectangular prism of given dimension.Since a rectangular prism whose base is 5 inches by 6 inchesThat is length (l)= 5 inches Width (w) = 6 inches and height (h) = 5 inches.Also . Substitute, we have,simplify , we have,surface area of a rectangular prism = 170 inches²Thus, the surface area of a rectangular prism of given dimension is 170 square inches

8 0

3 years ago

The diagram shows a prism of length 90 cm.

Xelga [282]

Answer:

xmmxxnxnnxnxnsnsnsnsnsnsnsndndbd

6 0

3 years ago