All categories

1 year ago

No links or wrong answers please!! 55 points

Mathematics

1 answer:

juin [17]1 year ago

4 0

The areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)

<h3>How to determine the total areas?</h3>

The figure 1

In this figure, we have

Length = x + 1

Width = 4

The area is calculated as:

Area = Length * Width

So, we have

Area = 4(x + 1)

The figure 2

In this figure, we have

Length = d + 4

Width = 7

The area is calculated as:

Area = Length * Width

So, we have

Area = 7(d + 4)

The figure 3

In this figure, we have

Length = y + 3

Width = y

The area is calculated as:

Area = Length * Width

So, we have

Area = y(y + 3)

Hence, the areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)

You might be interested in

5 divided by 5+4 x 12

AleksandrR [38]

Answer:

49 :))

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Help me pleaseeeeeeeeee

matrenka [14]

The answer would be D. Because the solution is where the lines intersect.
They only intersect at the origin in D.

5 0

2 years ago

18,932,612 – 9,096,089 =

alina1380 [7]

Answer:

9836523

is the answer. just check

4 0

3 years ago

Evaluate the expression w^2 - v + 1 for w = -2 and v = -8. -3 -11 13 5

seraphim [82]

-3
Because -2^2-8+1
-2to the second power is 4
4-8=-4
-4+1=-3

8 0

2 years ago

Find lim x→3 sqrt 2x+3-sqrt 3x/ x^2-3x. you must show your work or explain your work in words plsss I need help

sineoko [7]

I'm assuming the limit is supposed to be

$\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}$

Multiply the numerator by its conjugate, and do the same with the denominator:

$\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)$

so that in the limit, we have

$\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

Factorize the first term in the denominator as

$x^2-3x=x(x-3)$

The $x-3$ terms cancel, leaving you with

$\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

and the limand is continuous at $x=3$ , so we can substitute it to find the limit has a value of -1/18.

7 0

2 years ago