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3 years ago

Which ordered pair is on the graph of the function? (-6, 4) (-1, 2) (-2, -4) (6, -3)

Mathematics

2 answers:

VikaD [51]3 years ago

8 0

Answer:

(-6,4) hope that helps you it's that in the y there is two 4

ASHA 777 [7]3 years ago

6 0

The answer
Is :(-6,4)

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Find the value of x<br><br> (look at screenshot below)

Black_prince [1.1K]

Answer:

x = 25

Step-by-step explanation:

120 = 70 + 2x

120 - 70 = (70 - 70) + 2x

50 = 2x

50/2 = 2x/2

x = 25

5 0

3 years ago

Read 2 more answers

Autumn bought 4 pairs of jeans for her kids and one shirt for herself. The shirt cost $16.99. She spent a total of $108.95. What

klio [65]

Answer:

Each Jean cost $22.99

Step-by-step explanation:

First you would subtract the cost of the shirt from the total.

108.95-16.99=91.96

Then you would divide the 91.96 by 4 to get the cost for each Jean.

91.96/4=22.99

j=$22.99

6 0

3 years ago

two verticies of paralellogram ABCD are A(-6,-1) and B(-5,3). The intersection of the diagonals, E, are (-2, 1/2). Determine the

masya89 [10]

Answer:

D(2,2)

Step-by-step explanation:

The diagonals of a parallelogram gram bisects each other.

Therefore E is the midpoint of AD.

Let the coordinates of D be (a,b).

By the midpoint rule:

$( \frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}) = ( - 2, \frac{1}{2} )$

This implies that:

$( \frac{ - 6+a}{2} ,\frac{ - 1+b}{2}) = ( - 2, \frac{1}{2} )$

This implies that:

$( \frac{ - 6+a}{2} = - 2 ,\frac{ - 1+b}{2} = \frac{1}{2} )$

$( - 6+a =- 4 , - 1+b = 1 ) \\ ( a =- 4 + 6 , b = 1 + 1 ) \\ ( a =2 , b =2 )$

7 0

3 years ago

What is the surface area of this triangular pyramid? 3 m 3 m 3 m 3 m 2.6 m

tekilochka [14]

$\stackrel{\textit{3 standing triangles}}{3\left[\cfrac{1}{2}(\stackrel{b}{3})(\stackrel{h}{3}) \right]}~~ + ~~\stackrel{\textit{triangular base}}{\cfrac{1}{2}(3)(2.6)}\implies 13.5~~ + ~~3.9\implies 23.4~m^2$

4 0

3 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

3 years ago