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3 years ago

I need major help on this question try your best

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

6 0

First find sq 17 and sq 12
sq 17 = 4.123
sq 12 = 3.464
Now you can list them from greatest to least
5.03709...
4.923935
sq 17
sq 12

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The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the e

babymother [125]

The answer is A. To understand that, you need to put it in order by the x's and the y's to get x^2-6x + y^2-16y = -48. Now complete the square on both the x and the y terms to get (x^2-6x+9) +(y^2-16y+64) = -48+9+64. Rewriting that in vertex form on the left and doing the math on the right gives you
(x-3)^2 + (y-8)^2 = 25, which shows you a center of (3,8) and a radius of 5.

8 0

3 years ago

Read 2 more answers

Write two properties of right angled triangle

leonid [27]

Answer:

The properties of right angle triangle are as follows

the hypotenuse is the longest side
one angle is 90 degree then other angle will be also 90 degree.

5 0

2 years ago

I DONT KNOW THESE! HELP!

kkurt [141]

Answer:

20. x = 5, y = -2

21. x = 11, y = 12

22. x = 22, y = 11

23. x = 11, y = 10

Step-by-step explanation:

20. Opposite sides in the parallelogram are congruent, so

$2y+18=3x-1\\ \\6x-3=17-5y$

Solve this system of two equations:

$\left\{\begin{array}{l}2y-3x=-19\\ \\6x+5y=20\end{array}\right.$

Multiply the first equation by 2 and add two equations:

$2(2y-3x)+6x+5y=2\cdot (-19)+20\\ \\4y-6x+6x+5y=-38+20\\ \\9y=-18\\ \\y=-2$

Substitute it into the first equation:

$2\cdot (-2)-3x=-19\\ \\-3x=-19+4\\ \\-3x=-15\\ \\x=5$

21. Opposite angles in the parallelogram are congruent, so

$11x+5=10y+6$

Consecutive angles are supplementary, so

$6x-y+11x+5=180^{\circ}$

Solve this system of two equations:

$\left\{\begin{array}{l}11x-10y=1\\ \\17x-y=175\end{array}\right.$

From the second equation

$y=17x-175$

Substitute it into the first equation:

$11x-10(17x-175)=1\\ \\11x-170x+1750=1\\ \\-159x=-1749\\ \\x=11\\ \\y=17\cdot 11-175=187-175=12$

22. Opposite angles in the parallelogram are congruent, so

$2x-5=3y-12$

Consecutive angles are supplementary, so

$2x-5+7y+x=180^{\circ}$

Solve this system of two equations:

$\left\{\begin{array}{l}2x-3y=-7\\ \\3x+7y=185\end{array}\right.$

From the first equation

$x=-3.5+1.5y$

Substitute it into the second equation:

$3(-3.5+1.5y)+7y=185\\ \\-10.5+4.5y+7y=185\\ \\-105+45y+70y=1,850\\ \\115y=1,850+105\\ \\115y=1,955\\ \\y=17\\ \\x=-3.5+1.5\cdot 17=22$

23. Opposite sides in the parallelogram are congruent, so

$2x+9=4y-9\\ \\3x-5=2y+8$

Solve this system of two equations:

$\left\{\begin{array}{l}2x-4y=-18\\ \\3x-2y=13\end{array}\right.$

Multiply the second equation by 2 and subtract it from the first equation:

$2x-4y-2(3x-2y)=-18-2\cdot 13\\ \\2x-4y-6x+4y=-18-26\\ \\-4x=-44\\ \\x=11$

Substitute it into the first equation:

$2\cdot 11-4y=-18\\ \\-4y=-18-22\\ \\-4y=-40\\ \\y=10$

3 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

The area of a mirror is 225 square inches, and its width is 13 3/4 inches. Will the mirror fit in a space that is 15 inches by 1

gavmur [86]

If you multiply length times width to get area, it's the same with division. You do area divided by length or width. The equation would be 225 divided by 13 3/4. 225/13 3/4= 16.36 inches. 16.36 > 16 inches, so the answer would be no. (The mirrors length and width are 13 3/4 inches by 16 9/25)

7 0

3 years ago