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3 years ago

Between which two integers does the square root of 17 fall between?

1 answer:

4 0

Two integers mean numbers that do not have decimals. Whole numbers

to figure out between what two integers get a calculator and do square root 17
which is 4.123 (to three decimal places)

thus it falls between 4 and 5. since it is above 4 and below 5 and there are no integers in between possible.

Hope this helps :P

You might be interested in

Given vectors A (2,-1,5), B (4,3,-2) & C (5,4,0), find:?

MrRissso [65]

A) 4a - 3b + 2c

4(2, -1, 5) - 3(4, 3 , -2) + 2(5, 4, 0) = (8, -4, 20) - (12, 9, - 6) + (10, 8, 0) =

= (8 - 12 + 10 , -4 - 9 + 8 , 20 + 6 + 0) = (6, - 5, 26)

Answer: (6, - 5, 26)

b) magnitude of vector b

$\sqrt{4^2+3^2+(-2)^2} = \sqrt{16+9+4} = \sqrt{29} ~ 5.4$

c) vector of length 7 parallel to vector c

=> m(5,4,0) = (5m,4m,0)

=> $\sqrt{(5m)^2+(4m)^2+0}= \sqrt{25m^2+16m^2}= \sqrt{41m^2}=m \sqrt{41}=7$

=> m = 7 / √41 ≈ 1.093

=> 1.093 (5, 4, 0) = (5.465 , 4.372, 0)

Answer: (5.465 , 4.372 , 0)

7 0

3 years ago

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17 eq(1)

y = -1/2x eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 5/4, b =0 and c = -17

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

System A has 4 real solutions.

System B

y = x^2 -7x + 10 eq(1)

y = -6x + 5 eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-1 and c =5

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9 eq(1)

8x - y = -17 eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17

8(-2) - y = -17

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0

3 years ago

Read 2 more answers

Solve. 4 – d < 4 + d d > 8 d > 0 d > –8 d > –4

yaroslaw [1]

Answer:

-2d + 4 < 4

-2d < 0

d > 0

answer is b

3 0

3 years ago

HI HELP ITS A QUIZZZ!

Virty [35]

B. Its suggesting something bigger than -5

4 0

3 years ago

A little stuck on this problem help

KatRina [158]

Answer:

∠N = 31

Step-by-step explanation:

MN = MP

∠N = ∠P = x+3

∠M + ∠N + ∠P = 180

4x+6 + x+3 + x+3 = 6x + 12 = 180

x = 28

∠N = 28 + 3 = 31

5 0

3 years ago