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4 years ago

Write an equation in slope-intercept form for a line that passes through points (-1, 2) and (5, -10)

Mathematics

1 answer:

sesenic [268]4 years ago

6 0

The equation is y= -12/6x
So negative 12 over 6 x^^

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II The small capillaries in the lungs are in close contact with the alveoli. A red blood cell takes up oxygen during the 0.5s th

dangina [55]

The diffusion time of the red blood cell to take oxygen through the membrane is 25uS.

The small capillaries in the lungs are in close contact and it takes a red blood cell 0.5 seconds to squeeze through the capillary.

The diffusion time for the oxygen across the 1 um thick membrane separating the air is given by,

t = x²/2D

Where,

t is the diffusion time,

x is the thickness of the membrane,

D is the diffusion coefficient.

Putting values,

t = (1 x 10⁻⁹)²/2 x 2 x 10⁻¹¹

t = 10⁻⁷/4

t = 25 x 10⁻⁹ seconds.

t = 25 uS.

so, the diffusion time is 25uS.

To know more about Diffusion, visit,

brainly.com/question/94094

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3 0

2 years ago

Which expression is equivalent to 10 of the power of 5 ? 10·5 510 5·5·5·5·5·5·5·5·5·5 10·10·10·10·10

stira [4]

Answer is
10.10.10.10.10

8 0

3 years ago

Read 2 more answers

Y= -3x^2-24x-46 write in vertex form

mario62 [17]

ANSWER

$y= -3{(x + 4)}^{2} + 2$

EXPLANATION

The given function is:

$y= -3x^2-24x-46$

Factor -3 for the first 2 terms.

$y= -3(x^2 + 8x)-46$

We add and subtract the square of the coefficient of x.

$y= -3(x^2 + 8x + {(4)}^{2} ) - - 3( {4)}^{2} -46$

Factor the perfect square trinomial.

$y= -3 {(x + 4)}^{2} ) + 48 -46$

The vertex form is:

$y= -3{(x + 4)}^{2} + 2$

6 0

4 years ago

A number is 42,300 when multiplied by 10 Find the product of this and 500?

Simora [160]

2115000

x*10= 42300

x= 4230

x*500

4230*500= 2115000

Hope this helps

4 0

3 years ago

Read 2 more answers

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

3 years ago