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3 years ago

Graph y > x^2 - 5. Click on the graph until the correct one appears.

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

$y=x^2\\\\for\ x=\pm2\to y=(\pm2)^2=4\to(-2;\ 4);\ (2;\ 4)\\\\for\ x=\pm1\to y=(\pm1)^2=1\to (-1;\ 1);\ (1;\ 1)\\\\for\ x=0\to y=0^2=0\to (0;\ 0)$

Shift the graph of the function y = x², 5 units down /look at the picture #1/.

$y > x^2-5$ /look at the picture #2/ - your answer

ehidna [41]3 years ago

7 0

Answer:

Graph 2

Step-by-step explanation:

Here, the given inequality is,

$y>x^2-5-----(1)$

At (0,0),

$0>0-5$

$\implies 0 > -5$ ( True )

So, The shaded region must contain the origin,

Hence, Graph 1 and Graph 3 can not be the graph of $y>x^2-5$ ,

Now, we know that, The standard form of a parabola is,

$y=a(x-h)^2+k$

Where, (h,k) is the vertex of the parabola,

From equation (1), the related equation of inequality (1) can be written,

$y=(x-0)^2+(-5)$

By Comparing,

The vertex of the related parabola of inequality (1) is (0,-5),

⇒ The vertex of the given parabola must be lie on the negative y-axis

Thus, Graph 3 can not be the graph of the given equation,

Therefore, Graph 2 must be the graph of the given inequality.

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Cedric purchased a new refrigerator and stove at Home Depot for $3,729 on a deferred payment plan with no down

lesya [120]

Answer:

correct option is d. $242.81

Step-by-step explanation:

given data

APR = 25.5% = $\frac{25.5}{12}$ = 2.125

paid = $3,729

solution

we get here finance charge on the 1st month by multiplying 3,729 and now adding it to existing balance

so we get finance charge for the second and third months similarly as

APR ÷ 100 = $\frac{2.125 }{100}$ = 0.02125

so 1st

= $3,729 × 0.02125

= 79.25

and

$3,729 + $79.25 = $3808.24

so for next

= $3808.24 × 0.02125

= 80.93

and

$3808.24 + $80.93 = $3889.17

so for next

= $3889.17 × 0.02125

= 82.64

and

$3889.17 + $ 82.64 = $3971.81

finance charge = 3971.81 - 3729

finance charge = 242.81

so correct option is d. $242.81

8 0

3 years ago

Determine the truth value of each of these statements if thedomainofeachvariableconsistsofallrealnumbers.

hoa [83]

Answer:

a)TRUE

b)FALSE

c)TRUE

d)FALSE

e)TRUE

f)TRUE

g)TRUE

h)FALSE

i)FALSE

j)TRUE

Step-by-step explanation:

a) For every x there is y such that $x^2=y$ :

TRUE

This statement is true, because for every real number there is a square number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.

b) For every x there is y such that $x=y^2$ :

FALSE

For example, if x = -1, there is no such real number so that its square equals -1.

c) There is x for every y such that xy = 0

TRUE

If we put x = 0, then for every y it will be xy=0*y=0

d)There are x and y such that $x+y\neq y+x$

FALSE

There are no such numbers. If we rewrite the equation we obtain an incorrect statement:

$x+y \neq y+x\\x+y - y-y\neq 0\\0\neq 0$

e)For every x, if $x \neq 0$ there is y such that xy=1:

TRUE

The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.

f)There is x for every y such that if $y\neq 0$ then xy=1.

TRUE

The statement is equivalent to the statement in e)

g)For every x there is y such that x+y = 1

TRUE

The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x

So, the statement says that for every real umber there is a real number that is equal to 1-that number

h) There are x and y such that

$x+2y = 2\\2x+4y = 5$

We have to solve this system of equations.

From the first equation it yields x=2-2y and inserting that into the second equation we have:

$2(2-2y)+4y=5\\4-4y+4y=5\\4=5$

Which is obviously false statement, so there are no such x and y that satisfy the equations.

FALSE

i)For every x there is y such that

$x+y=2\\2x-y=1$

We have to solve this system of equations.

From the first equation it yields $x=2-y$ and inserting that into the second equation we obtain:

$2(2-y)-y=1\\4-2y-y=1\\4-3y=1\\-3y=1-4\\-3y=-3\\y=1$

Inserting that back to the first equation we obtain

$x=2-1\\x=1$

So, there is an unique solution to this equations:

x=1 and y=1

The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that

$x+y=2\\2x-y=1$

j)For every x and y there is a z such that

$z=\frac{x+y}{2}$

TRUE

The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.

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3 years ago

What equation of the line that has a slope of -4 and passes through the point (-2,-5)? Answer ASAP!!

Anna11 [10]

Answer:

Given points are ( -2 , -5 ) and slope ( m) = -4

Step-by-step explanation:

y - y1 = m ( x - x1)

y + 5 = -4 ( x + 2 )

y + 5 = -4x - 8

4x + y +5 + 8 = 0

4x + y +13 = 0 

y = -4x - 13

which is the required equation 

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3 years ago

What does the "area" of a pizza mean?

denis23 [38]

Answer:

The area is the entire pizza

Step-by-step explanation:

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2 years ago

You offer senior citizens a 20% discount on their tune-ups at your gas station. Assuming that an average of 50 senior citizens g

GREYUIT [131]

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3 years ago