All categories

3 years ago

Hello anyone can do this? Please put answer in points(x,y)

Mathematics

1 answer:

zimovet [89]3 years ago

3 0

Answer:

Check the attached graph

Step-by-step explanation:

Given that slope of the line is m = 1

And the line passes through the point (-3,5).

Now we need to graph the line using above information. So begin by graphing the point (-3,5).

Slope means rise/run

Then slope = 1 means rise/run=1=1/1

means move point 1 unit up then 1 unit right

then new point is at (-2,6).

Now draw a line joining both points to get the final graph.

You might be interested in

Which of the following is a counterexample for this conditional statement?If an appliance is used for cooling, then it is a refr

Nezavi [6.7K]

Explanation

We are asked to give a counter-example for the given statement

A counter-example is an example that opposes or contradicts an idea or statement.

Therefore, for the statement

we have that

Statement: If an appliance is used for cooling

Conclusion: then it is a refrigerator

Then, the option that contradicts this, will be a clothes dryer

Because a dryer doesn't cool

Hence, the answer is a cloth dryer

option D

4 0

1 year ago

What is the shape of the horizontal cross section of the

GuDViN [60]

Answer: Pentagon

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Which is the graph of f(x)=1/4(4)x

Dovator [93]

Answer:

Please, see attached picture

Step-by-step explanation:

We can easily solve this question by using a graphing calculator or any plotting tool.

The equation is

f(x) = (1/4)*(4)^(x)

Which is an exponential function, that grows rapidly to infinity as x increases.

8 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

What is the next number in the pattern? 7,11,10,14,13

nikdorinn [45]

I think the answer is 17 because the pattern adds 4 then takes away 1 so 14-1=13 13+4=17

6 0

3 years ago