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2 years ago

Which statement is NOT true about the graph of y=5x?

Mathematics

1 answer:

777dan777 [17]2 years ago

8 0

Answer:

C) The graph intersects the y-axis at y=1.

Step-by-step explanation:

C is untrue because the y-intercept is given by the term "b" in the equation:

$y = mx+b$

Here, m = 5 and b is equal to 0, this the graph intersects the y-axis at 0 not 1. So, C is untrue.

You might be interested in

Find the radius and the diameter to the nearest hundredth when C = 7π yd. Use 3.14 for π.

valentinak56 [21]

Answer:

Diameter = 7

Radius = 3.5

Step-by-step explanation:

Hi! Ace here.

So, the formula to find circumference is dπ or 2rπ

So to find the diameter, you do the circumference/π

To find the radius, you'd do the circumference/2π

Plug-in those numbers and there you go!

Hope this helps, let me know if you have any questions.

I know this was a very vague explanation so I encourage you to ask any questions. Thank you!

5 0

2 years ago

Three friends decided to exercise together four times a week to lose fatand increase muscle mass. While all three were healthier

gregori [183]

Answer:

6 pounds lost

Step-by-step explanation:

5+4 is 9 9-3 is 6 which means that in total, 6 pounds was lost

5 0

2 years ago

Is y= 3(1/2)^x increasing or decreasing?

Zigmanuir [339]

Answer: Hey I looked on a graphing calculator on Desmos, and its seems that this is increasing.

Step-by-step explanation:

5 0

2 years ago

Can someone help with this ??

Zigmanuir [339]

I’m sorry but I can’t see the photo so can you tape it out so I can furthermore assist you

6 0

2 years ago

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f

Korvikt [17]

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$

5 0

2 years ago