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3 years ago

Given the equation y = 5x - 1, what are the intercepts?

Mathematics

2 answers:

IceJOKER [234]3 years ago

6 0

The x intercept would be 1/5, the y intercept would be -1

andriy [413]3 years ago

4 0

Answer:

x-intercept = 1/5 and y-intercept = -1

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-6x-7=4x-2. Can you show step by step explanation. thank u.

monitta

Answer:

x=-1/2

Step-by-step explanation:

-6x-7=4x-2

-6x-7+6x=4x+6x-2

-7=10x-2

-7+2=10x-2+2

-5=10x

-5/10=10x/10

x=-1/2

Hope this helps!

8 0

2 years ago

find the coordinates of the circumcenter of triangle abc. A(0,5) B(-4,5) C(-4,-3) PLEASE HELP IM FAILING AND I WILL GIVE BRAINLI

hodyreva [135]

(-2,1). Hope this helps!

4 0

3 years ago

What is (−2.1)⋅(−1.4)

professor190 [17]

Multiplying a negative number and another negative number makes the product positive.
So (-2.1)*(-1.4) = 2.94

8 0

1 year ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Which is the graph of the sequence defined by the function f(x + 1) = 2/3 f(x) if the initial value of the sequence is 108?

san4es73 [151]

Answer:

The fourth graph.

Step-by-step explanation:

We have f(x + 1) so the x values will be increased by 1.

first value = 108 (when x = 1)

next value = 108 * 2/3 = 72 (when x = 2)

next = 72 * 2/ 3= 48 (when x = 3)

next = 48 * 2/3 = 32 (when x = 4)

5 0

3 years ago

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