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Lady bird [3.3K]
2 years ago
6

Please answer this if you can! <3 (NO EXPLANATION REQUIRED!)

Mathematics
1 answer:
denpristay [2]2 years ago
4 0
Answer: 5 bigger or equal but less than 10 the frequency is 2.
10 bigger or equal but less than 15 the frequency is 6.
15 bigger or equal but less than 20 the frequency is 10.
20 bigger or equal but less than 25 the frequency is 2.
And if u add all of the frequency it’s 20.
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What's the answer for 24
scZoUnD [109]
X= -2 or X= 3 either one works.
3 0
3 years ago
John made a rectangle pen for his dog using 28 feet of fencing. If the width of the pen is 2 feet more than one-half the length,
kati45 [8]
Okay, so you know that the perimeter of the pen is 28 feet. Now we need the separate side lengths.

P = 28 ft.
Length = x
Width = (1/2)x + 2

So now we can plug in the little equations of length and width into a perimeter formula:
 2(x) + 2((1/2)x + 2) = 28
Distribute the 2's and solve as needed.
2x + x + 4 = 28
3x + 4 = 28
3x = 24
x = 8

To find the width, just plug in 8 for x in our little equation:
(1/2)(8) + 2
= 6

So, the length is 8 feet and the width is 6 feet.
5 0
3 years ago
Read 2 more answers
Cara evaluated the expression below. (4(7-13))÷3+(-4)²-2(6-2) =(28-13)÷3+(-4)²-2(4) =15÷3+16-8 =5+16-8 =13 What was Cara’s error
mezya [45]

Answer:

the answer is the first one, Cara multiplied only 4 and 7 together and did not multiply 4 and 13

Step-by-step explanation:

if you look through the question you can see that Cara evaluated the (-4)^2 correctly because a negative multiplied by a negative gives you a positive thus (-4)*(-4)= 16, then next you can see that Cara subtracted 2 from 6 correctly because (6-2)= (4), and last but not least Cara did multiply 2 by 4 correctly because 2*4 or 2(4) =8 so as you can see the answer is the first one

3 0
3 years ago
Read 2 more answers
What is the domain and range of the function y = 2x2 - 4x - 10?
11111nata11111 [884]

Answer:

Step-by-step explanation:

The domain of all polynomials is all real numbers.  To find the range, let's solve that quadratic for its vertex.  We will do this by completing the square.  To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out.  That gives us:

2(x^2-2x)=10

The second rule is to take half the linear term, square it, and add it to both sides.  Our linear term is 2 (from the -2x).  Half of 2 is 1, and 1 squared is 1.  So we add 1 into the parenthesis on the left.  BUT we cannot ignore the 2 sitting out front of the parenthesis.  It is a multiplier.  That means that we didn't just add in a 1, we added in a 2 * 1 = 2.  So we add 2 to the right as well, giving us now:

2(x^2-2x+1)=10+2

The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form.  Completing the square gives us a perfect square binomial on the left.

x^2-2x+1=(x-1)^2 and on the right we will just add 10 and 2:

2(x-1)^2=12

Now we move the 12 back over by subtracting and set the quadratic back to equal y:

2(x-1)^2-12=y

From this vertex form we can see that the vertex of the parabola sits at (1,-12).  This tells us that the absolute lowest point of the parabola (since it is positive it opens upwards) is -12.  Therefore, the range is R={y|y ≥ -12}

4 0
3 years ago
How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%
blondinia [14]

The trick is to exploit the difference of squares formula,

a^2-b^2=(a-b)(a+b)

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2

Whatever you do to the denominator, you have to do to the numerator too. So

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2

Expand the numerator:

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)

(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}

So we have

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2

But √12 = √(3•4) = 2√3, so

\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}

7 0
2 years ago
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