HELPPP PLSSSSSSSSSSSSS

Mathematics

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

x=(-1/10) is the answer of given algebra

You might be interested in

I need the answers for 3, 4, 5, and 6! thanks!!

Makovka662 [10]

Number 3 =32 us cups

4 0

3 years ago

Dana is growing vegetables in planters. She has 8 bags of soil and he uses 2/3 of the bag of soil to fill each planter. How many

Korolek [52]

Dana can fill 12 planters. Let me know if you want an explanation!

4 0

3 years ago

Read 2 more answers

If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3

horrorfan [7]

Answer: The correct option is (B) 24 : 25.

Step-by-step explanation: Given that the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2 : 3.

We are to find the ratio of the area of R to the area of S.

Let 2x, 3x be the sides of rectangle R and y be the side of square S.

Then, according to the given information, we have

$\textup{Perimeter of rectangle R}=\textup{Perimeter of square S}\\\\\Rightarrow2(2x+3x)=2(y+y)\\\\\Rightarrow 5x=2y\\\\\Rightarrow \dfrac{x}{y}=\dfrac{2}{5}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Therefore, the ratio of the area of R to the area of S is

$\dfrac{2x\times3x}{y\times y}\\\\\\=\dfrac{5x^2}{y^2}\\\\\\=6\left(\dfrac{x}{y}\right)^2\\\\\\=6\times\left(\dfrac{2}{5}\right)^2~~~~~~~~~~~[\textup{Using equation (i)}]\\\\\\=\dfrac{24}{25}\\\\=24:25.$

Thus, the required ratio of the area of R to the area of S is 24 : 25.

Option (B) is CORRECT.

6 0

3 years ago

What is the mesure of

lara [203]

What are we measuring

4 0

3 years ago

Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

7 0

3 years ago