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2 years ago

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Could someone please help

1 answer:

alexdok [17]2 years ago

3 0

Answer:

Answer:<em>I think that is right...</em>

Answer:<em>I think that is right...Please make me brainliest...</em>

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(08.01)Two lines, A and B, are represented by the following equations:

algol [13]

Answer:

The second one :) Hope it helps

Step-by-step explanation:

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3 years ago

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Please answer the following 4 questions

Lilit [14]

1. The correct answer should be A
2. The answer should be C
3. I think the answer is D
4. The answer should be D

Hope this helps :)

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3 years ago

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What is the slope-intercept form equation of the line that passes through (1, 3) and (3, 7)? (1 point) y = −2x + 1 y = −2x − 1 y

Stolb23 [73]

Answer: y=2x+1

Step-by-step explanation:

The slope of the line is $\frac{7-3}{3-1}=\frac{4}{2}=2$

Using the point (1,3) to substitute into point-slope form,

$y-3=2(x-1)\\\\y-3=2x-2\\\\\boxed{y-2x+1}$

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2 years ago

if a quadratic equation with real coefficients has a discriminant of -36 then what type of roots does it have

Aleks04 [339]

Step-by-step explanation:

We have,

If a quadratic equation with real coefficients has a discriminant of -36.

The general form of quadratic equation is :

$ax^2+bx+c=0$

The discriminant of this equation is : $D=b^2-4ac$

If D=0, it will have 1 real roots

If D>0, it will have 2 real roots

If D<0, it will have no real roots

We have,

D = -36 < 0, so, the quadratic equation will have no real roots.

5 0

3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

3 years ago