All categories

3 years ago

Area for rectangle in the coordinate plane. IXL Geometry help pls !

Mathematics

1 answer:

Sidana [21]3 years ago

5 0

Answer:

Step-by-step explanation:

The co-ordinates of the rectangle are:

A (8, 6)

B (-4, -10)

C (-8, -7)

D (4, 9)

The Area of Rectangle is Given by: length x breadth = AB x BC

Lengths of AB and BC can be found by distance formula

length(AB) = $\sqrt{(x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2}}$ = $\sqrt{(-4-8)^{2} + (-10-6)^{2}} = 20$

Similarly,

length(BC) = 5

Area of Rectangle = AB x BC = 20 x 5 = 100 square units

You might be interested in

Assuming that the Earth has a diameter of 7920 miles and a volume 6.7 times that of the planet Mars, what is the diameter of Mar

pochemuha

53,064 hope this helps
god bless

6 0

3 years ago

Show twelve in four different ways

nlexa [21]

Ok i hope this is the answer that you are looking for
6*2=12one way
3×4=12second way
12×1=12third way
Sorry if this is not what you are looking for and I couldn't find the fourth way!!
Or incase there is another ways like 6+6=12,9+3=12,12+0=12,24÷2=12
Sorry if I did this wrong

4 0

3 years ago

Mrs. whimsey's class is having a pizza party . If there are 16 student and 4 pizzas cut into 8 slices , what answer choice repre

777dan777 [17]

each student recieves 2 slices of pizza because 4 pizzas cut into 8 slices divided among 16 students ot 4x8/16=2. 4x8=32 32/16=2

3 0

3 years ago

Which one A or B or C help pls!

Marrrta [24]

Answer:

i think it c

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Ex 4.8 14) integrate 3^(2x-1)

sesenic [268]

Take the integral:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{1}{2}\cdot 2\cdot 3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^{2x-1}\cdot 2\,dx\qquad(i)} \end{array}$

$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{2x-1=u\quad\Rightarrow\quad 2\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^u\,du}\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{1}{\ell n\,3}\,3^u+C}\\\\ =\mathsf{\dfrac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!3^{2x-1}\,dx=\frac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}}\qquad\checkmark \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2154165

$\large\textsf{I hope it helps. :-)}$


Tags: integrate indefinite integral substitution exponential base logarithm log ln composite integral calculus

4 0

3 years ago