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sammy [17]
4 years ago
13

Please show your work

Mathematics
1 answer:
Neporo4naja [7]4 years ago
5 0

Answer:

it is the butt-tox y in this equation and my peepee hole says it smells in the graph so nice poopy job random person!!!!!!!????

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Explain the steps how u got it<br> •••••••••••••••••••••••••••••
harina [27]

Answer:

the answer would be c

Step-by-step explanation:

≤ less than or equal to sign

2by some number take away 4 is less than or equal to 26

4 0
3 years ago
Read 2 more answers
Determine whether the points p and q lie on the given surface. r(u, v) = u + v, u2 − v, u + v2 p(3, 3, 3), q(4, −2, 10)
Troyanec [42]

Answer:

If  p  and q  lie on the surface , then we should be able to find values of  u  and  v  such that they can satisfy the equation of the surface.

i.e

x = u+v;  

y = u^2 - v and

z = u + v^2

For p we have to find u and v such that:

u + v =3 ;                      ......[1]

u^2 - v = 3 ;     .....[2]

u + v^2 = 3       .....[3]

Equate [1] and [3] we have;

u + v = u + v^2

Subtract u from both sides we get;

u + v -u= u + v^2-u

Simplify:

v = v^2

or

v^2 - v = v(v-1) = 0

⇒ v = 0 and v = 1

To find the value of u substitute the values of v in [1]

for v =0

u + 0 = 3

u = 3

for v = 1

u + 1 = 3

Subtract 3 from both sides we get;

u =  2

Now; substitute the values of u and v in equation [2] to satisfy:

if u = 3 ad v = 0

then;

u^2 - v = 3

3^2 - 0 = 3

9 = 3 which is not true

if u =2 and v =1

then;

2^2 - 1 = 3

4-1 = 3

3 = 3 which is true.

Therefore, the point p lies on the given surface with u = 2 and v =1

Similarly, for q we have to find u and v such that:

u + v =4 ;                      ......[1]

u^2 - v = -2 ;     .....[2]

u + v^2 = 10       .....[3]

Adding first two equation; we get

u + v +u^2 -v = 4-2

Simplify:

u^2+u -2 =0

or

(u+2)(u-1) = 0

⇒ u = -2 and u = 1

from [1] we have  v = 4- u

For u = -2

then;

v = 4-(-2) = 4+2

v = 6

For u = 1

then;

v = 4-1

v =3

Now; substitute the values of u and v in equation [3] to satisfy:

if u = -2 ad v = 6

then;

u +v^2 = 10

-2 +6^2 = 10

-2 +36 = 10

34 = 10 which is not true

if u = 1 ad v = 3

then;

u +v^2 = 10

1 +3^2 = 10

1 +9 = 10

10 = 10 which is not true.

Therefore, the point q lies on the given surface with u=1 and v =3

5 0
3 years ago
Solve the system of equations by graphing. <br><br> x+y=3<br> y=2x−15
Stella [2.4K]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x + y = 3

x = 3 - y


y = 2x - 15

y = 2(3 - y) - 15

y = 6 - 2y - 15

y = -2y - 9

3y = -9

y = -3


y = 2x - 15

-3 = 2x - 15

-2x = -12

x = 6


So...

The answer is: (6, -3)

4 0
4 years ago
Read 2 more answers
Determine which integers in the set S: {−2, −3, −4, −5} will make the inequality 2p − 8 &lt; 5p + 4 true.
Alika [10]

The set S:{ - 2 , - 3 } will make the inequality 2p - 8 < 5p + 4 true


What is an Inequality Equation?

Inequalities are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.

In an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, <, ≤ or ≥.

Given data ,

The set S : { - 2 , - 3 , - 4 , - 5 }

The inequality relation is 2p - 8 < 5p + 4

Now , substituting the value from set S in the inequality , we get

When p = - 2 :

2p − 8 < 5p + 4

2(-2) − 8 < 5(-2) + 4

-4 − 8 < -10 + 4

-12 < -6  TRUE

When p = - 3

2p − 8 < 5p + 4

2(-3) − 8 < 5(-3) + 4

-6 − 8 < -15 + 4

-14 < -11  TRUE

When p = - 4

2p − 8 < 5p + 4

2(-4) − 8 < 5(-4) + 4

-8 − 8 < -20 + 4

-16 < -16  FALSE

When p = - 5

2p − 8 < 5p + 4

2(-5) − 8 < 5(-5) + 4

-10 − 8 < -25 + 4

-18 < -21  FALSE

Therefore , the value of p = - 2 , - 3 will satisfy the given inequality relation

Hence , the set S:{ -2,-3 } is the solution

To learn more about inequality equation click :

brainly.com/question/11897796

#SPJ2

3 0
1 year ago
Read 2 more answers
I need help and please ahow your work
dusya [7]
There ya go! Hope this helped

7 0
4 years ago
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