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Anastasy [175]
3 years ago
6

15) Which equation does not represent a function?

Mathematics
1 answer:
givi [52]3 years ago
4 0
The answer is D as it can be rewritten as -((y^2)/5)-(8/5)=x. Solving for the x-intercepts you get -((y^2)/5)-(8/5) = 0 as you want to find the y values when x is zero. Solving for y you get: (y^2)/5)+(8/5) = 0 => (y^2)/5)=-(8/5) => y^2=-8 => y = plus or minus sqrt(-8). The first problem is that you have 2 x intercepts which already makes it not a function and second sqrt(-8) is an imaginary number making the solution not a real number.
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Can someone help me asap
djyliett [7]

Answer:

1) 4

2) 5

3) -23

4) 8

5) 7

6) -64

7) -10

8) 19

9) 9

10) -56

6 0
3 years ago
Read 2 more answers
The diagonals of a rectangle measure x+5 feet and 2x+1 feet what is the value of x?
vodomira [7]

The value of x is 4.

Given that diagonals of a rectangle measure x+5 feet and 2x+1 feet as shown in the attached figure.

The diagonal of a rectangle is a line or straight line that connects the opposite corners or vertices of the rectangle.

In the given figure ABCD is a rectangle.

OA=2x+1 and OD=x+5             [Given]

AC and BD are diagonals of a rectangle.

As we know that the diagonals of a rectangle are always equal.

So, AC = BD

We can also write it as,

2×OA=2×OD

2×(2x+1) =2×(x+5)

Apply the distributive property a(b+c)=ab+ac, we get

4x+2=2x+10

Subtract 2x from both sides, we get

4x+2-2x=2x+10-2x

2x+2=10

Subtract 2 from both sides, we get

2x+2-2=10-2

2x=8

Divide both sides by 2, we get

2x/x=8/2

x=4

Hence, the value of x=4 when diagonals of a rectangle measure x+5 feet and 2x+1 feet.

Learn more about rectangles from here brainly.com/question/1549055.

#SPJ4

4 0
2 years ago
12,10,15, 13.18.16.21.19 what comes next .<br><br><br><br><br><br>​
shtirl [24]
The answer is 24. it works like this. you subtract 2 from 12 and get 10. then add 5 to 10 to get 15. then subtract 2 from 15 and get 13. 13+5 is 18. and so on
3 0
2 years ago
Read 2 more answers
A small cube has the volume shown. Its side length is 1.5 in less than a second larger cube. What is
eduard

The volume of the larger cube is 227 inches cube.

<h3 /><h3 /><h3>Volume of a cube is describe below:</h3>
  • v = L³

where

L = length

The small cube has the following volume:

  • v = 95 in³

Let

The side length of the small cube = x

Therefore,

95 = x³

x = ∛95

x = 4.56290264

x ≈ 4.6

Its side length is 1.5 inches less than a second larger cube. Therefore,

  • length of larger length cube = (x + 1.5 )inches = 4.6 + 1.5 = 6.1 inches

Therefore, the volume of the larger cube can be found as follows:

Volume = 6.1³

volume = 226.981

volume ≈ 227 inches³

learn more on cubes here:brainly.com/question/2671825?referrer=searchResults

6 0
2 years ago
[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>​
asambeis [7]

Given:

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

To prove:

The given statement.

Proof:

We have,

\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)

LHS=\cos^4 \alpha+\sin^4\alpha

LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2

LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha     [\because a^2+b^2=(a+b)^2-2ab]

LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha      [\because \cos^2 \alpha+\sin^2\alpha=1]

LHS=1-2\cos^2 \alpha+2\cos^4 \alpha

Now,

RHS=\dfrac{1}{4}(3+\cos 4 \alpha)

RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]

RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]        [\because \cos 2\theta=2\cos^2\theta -1]

RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]        [\because (a-b)^2=a^2-2ab+b^2]

RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]

RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]

RHS=1+2\cos^4 \alpha-2\cos \alpha

RHS=1-2\cos^2 \alpha+2\cos^4 \alpha

LHS=RHS

Hence proved.

8 0
3 years ago
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