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julia-pushkina [17]

2 years ago

9

A rectangular garden is 16 feet long. If you walk diagonally across the garden, you would walk 20 feet. How many feet wide is th

e garden?

1 answer:

Ivan2 years ago

8 0

You can find the wide by using phytagoras dalil,
Wide =
$\sqrt{20 {}^{2} - 16 {}^{2} }$
The answer is 12 feet

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Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(2, 0)A(2,0)A

const2013 [10]

9514 1404 393

Answer:

28 square units

Step-by-step explanation:

The rectangle is 7-0 = 7 units high and 6-2 = 4 units wide. Its area is the product of these dimensions:

A = LW

A = (7)(4) = 28 . . . square units

3 0

2 years ago

Read 2 more answers

What are the x and y-intercepts of the line described by the equation?

tatuchka [14]

Answer: x=4.41 y= -7.35

Step-by-step explanation:

Pretty basic just graph them using Desmos...

Showing your work you would need to get x on one side then y on the other.

Because 3y is positive I personally like to use that one first. The inverse operation of 3y is divide so you divide 3y by 3 and the 3s cancel out. Divide -22.05 by 3 and you get -7.35. Which is your y value.

Finding X is a bit more difficult, you need to divide by -5 which is the obvious inverse operation. So, -22.05/-5 and you get 4.41. Which is your X value.

Any further questions feel free to contact me.

-Vy

5 0

2 years ago

A solid plastic ball is a sphere with a radius of 8 inches. How much plastic does it take to make one ball

mihalych1998 [28]

Answer:

2144.66in³

Step-by-step explanation:

Radius = 8 inches.

To determine the amount of sphere to make the ball, we'll have to find its volume.

Volume of a sphere = 4/3 * Πr³

V = 4/3*Πr³

Π = 22/7

V = 4/3 * 22/7 * 8³

V = 2144.66inches³

V = 2144.66in³

The amount of plastic required is 2144.66in³.

7 0

2 years ago

Which ordered pairs could be points on a line parallel to the line that contains (3, 4) and (–2, 2)? Check all that apply.

Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

As we go from (–2, 2) to (3, 4), x increases by 5 and y increases by 4. Thus, the slope of the line through (–2, 2) and (3, 4) is

m = rise / run = 4/5.

Use the slope-intercept form of the equation of a straight line:

y = mx + b becomes 4 = (4/5)(3) + b. Multiplying all three terms by 5, we eliminate the fraction: 20 = 12 + b. Thus, b = 8, and the equation of the line through (–2, 2) and (3, 4) is y = (4/5)x + 8.

A line parallel to this one would have the form y = (4/5)x + b; note that the slopes of these two lines are the same, but the y-intercept, b, would be different if the two lines do not coincide.

Unfortunately, you have not shared the ordered pairs given in this problem statement.

You could arbitrarily let b = 0. Then the parallel line has equation

y = (4/5)x; if x = 3, then y = (4/5)(3) = 12/5, and so (3, 12/5) lies on the parallel line.

3 0

2 years ago

Read 2 more answers

Prove that √2 +√5 is irrational

Sindrei [870]

We have to prove that $\sqrt{2}+\sqrt{5}$ is irrational. We can prove this statement by contradiction.

Let us assume that $\sqrt{2}+\sqrt{5}$ is a rational number. Therefore, we can express:

$a=\sqrt{2}+\sqrt{5}$

Let us represent this equation as:

$a-\sqrt{2}=\sqrt{5}$

Upon squaring both the sides:

$(a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}$

Since a has been assumed to be rational, therefore, $\frac{a^{2}-3}{2a}$ must as well be rational.

But we know that $\sqrt{2}$ is irrational, therefore, from equation $\sqrt{2}=\frac{a^{2}-3}{2a}$ the expression $\frac{a^{2}-3}{2a}$ must be irrational, which contradicts with our claim.

Therefore, by contradiction, $\sqrt{2}+\sqrt{5}$ is irrational.

4 0

3 years ago