27 milliliters .1/10 cm is the one and ten is the mm one =

Sin=95/100 you should Find an other edge if we say y is an other edge. so,100^2=95^2+y^2
Y=31.22
Cosx=31.22/100=
O.3122

6 0

3 years ago

The sum of two positive integers, x and y, is not more than 40. The difference of the two integers is at least 20. Chaneece choo

Butoxors [25]

Answer:

The value of x lies between 20 and 40 and value of y lies between -20 and 10

Step-by-step explanation:

We are given that

$y\leq 40-x$ ....(1)

$y\leq x-20$ ...(2)

First we convert inequality equation into equality equation to find the solution of the given system of inequality equation.

Therefore, we can write as

$y=-x+40$ ...(3)

$y=x-20$ ...(4)

Adding equation (3) and (4) we get

$2y=20$

$y=10$

Substitute y=10 in equation (3) we get

$10=40-x$

$x=40-10=30$

(30,10) is the intersect point of two equation.

Put x=0 in equation (3)

$y=40$

Substitute y=0 in equation (3)

$x=40$

Substitute x=0 in equation (4)

$y=-20$

Substitute y=0 in equation (4)

$x=20$

Substitute x=0 and y=40 and in equation (1)

$40\leq 40$

Hence, the equation is true.Therefore, the shaded region below the line.

Substitute x=0 and y=-20 in equation (2)

$-20\leq -20$

The equation is true. Hence, the shaded region is below the line.

Hence, the value of x lies between 20 and 40 and value of y lies between -20 and 10.

5 0

3 years ago

Read 2 more answers

The length of a parallelogram exceeds its breadth by 30 cm. If the perimeter of the parallelogram is 2 m 60 cm, find the length

Nuetrik [128]

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

$\longrightarrow \sf{2(1 + b) = 260}$

$\longrightarrow \sf{2(x+30+x) = 260}$

$\longrightarrow \sf{ 2x+30 = \dfrac{260}{20} }$

$\longrightarrow \sf{2x + 30 =130}$

$\longrightarrow \sf{2x = 130 - 30}$

$\longrightarrow \sf{2x = 100}$

$\longrightarrow \sf{x= \dfrac{100}{2} }$

• $\longrightarrow \sf{b= \underline{50cm}}$

$\longrightarrow \sf{= x + 30}$

$\longrightarrow \sf{= 50 + 30}$

• $\longrightarrow \sf{ \underline{80cm}}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\large\bm{Breadth= \: 50cm}$

$\large\bm{Length= \: 80cm}$

3 0

2 years ago

PLEASE HELP ITS DUE REALLY SOON!!!

oksian1 [2.3K]

Answer:

-(5x-21)/6 so d

Step-by-step explanation:

to do inverse functions you want to replace the x in the function with y so x=(21-6y)/5, then solve for y and you will get the answer I got and its d

8 0

3 years ago

here are several types of objects. For each type of object, estimate how many there are in a stack that is 5 feet high. Be prepa

lorasvet [3.4K]

We need to estimate how tall is, on average, one of these object, and then count how many would be in a 5 feet high stack.

For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

$1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}$

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

$0.75k=5 \iff k=6.\bar{6}$

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

$0.25k=5 \iff k=20$

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

$0.006k=5 \iff k=833.\bar{3}$

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0

3 years ago