Ask question

Ask question

All categories

2 years ago

13

PLEASE HELP PLEASE HELP

1 answer:

mars1129 [50]2 years ago

6 0

36 cuz formula for area is length times the width. 4 x 9 = 36

You might be interested in

Is this a function idk also what's the domain and range

masya89 [10]

This is a function since it passes the vertical line test. It is impossible to draw a single vertical line to have it pass through more than one point on the curve.

The domain is the set of numbers x such that x is between -3 and +3 excluding those endpoints. In other words, the domain is -3 < x < 3. We never actually get to either endpoint because of the vertical asymptotes.

The range is the set of all real numbers. It is possible to get any output we want depending on the specific input.

3 0

4 years ago

Please help me with A,B,c,D

g100num [7]

Answer:

wat is this, my g this looks like a test

7 0

3 years ago

a train is traveling from New Orleans to Memphis at a speed of 79 mph. New Orleans and Tennessee are 397.1 miles apart. How long

GrogVix [38]

<span>The train will need 5 hours to reach Tennessee</span>

7 0

3 years ago

Maria works as an electrician and earns $24.68/h. If she worked for 15 hours on one job, how much did she earn? *

rodikova [14]

Answer:

she would have $370.20

Step-by-step explanation:

24.68*15=370.2

8 0

3 years ago

50 PTS ANSWER ALL <3333333

11Alexandr11 [23.1K]

QUESTION 33

The length of the legs of the right triangle are given as,

6 centimeters and 8 centimeters.

The length of the hypotenuse can be found using the Pythagoras Theorem.

${h}^{2} = {6}^{2} + {8}^{2}$

${h}^{2} = 36+ 64$

${h}^{2} = 100$

$h = \sqrt{100}$

$h = 10cm$

Answer: C

QUESTION 34

The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.

The length of the other leg can be found using the Pythagoras Theorem,

${l}^{2} + {33}^{2} = {55}^{2}$

${l}^{2} = {55}^{2} - {33}^{2}$

${l}^{2} = 1936$

$l = \sqrt{1936}$

$l = 44cm$

Answer:B

QUESTION 35.

We want to find the distance between,

(2,-1) and (-1,3).

Recall the distance formula,

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

Substitute the values to get,

$d=\sqrt{( - 1-2)^2+(3- - 1)^2}$

$d=\sqrt{( - 3)^2+(4)^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d = 5$

Answer: 5 units.

QUESTION 36

We want to find the distance between,

(2,2) and (-3,-3).

We use the distance formula again,

$d=\sqrt{( - 3-2)^2+( - 3- 2)^2}$

$d=\sqrt{( - 5)^2+( - 5)^2}$

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}$

Answer: D

8 0

3 years ago

Read 2 more answers