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3 years ago

URGENT HELP PLZ

Mathematics

1 answer:

Aliun [14]3 years ago

8 0

Answer:

y = 1/3x + 4

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step 1: Find slope <em>m</em>

m = (2 - 3)/(-6 - -3)

m = -1/(-6 + 3)

m = -1/-3

m = 1/3

y = 1/3x + b

Step 2: Find y-intercept <em>b </em>(plug in coordinates)

3 = 1/3(-3) + b

3 = -1 + b

b = 4

Step 3: Rewrite linear equation

y = 1/3x + 4

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I NEED HELP ASAP!!!!

dedylja [7]

Answer:

B) 8

Step-by-step explanation:

Triangles MNC and RSC are similar because all 3 angles are the same, so the the ratio between each side should be the same.

If NC=12 and SC=6 then triangle MNC is scaled up by 2 from triangle RSC. This means each side of MNC is 2 times bigger than on RSC. So, sense RS=4, we can multiply that by 2 to get MN=8.

8 0

2 years ago

Please help meeeee math

Sergeeva-Olga [200]

Q2. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

We have

$x=2t \implies \dfrac{dx}{dt}=2$

$y=t^4+1 \implies \dfrac{dy}{dt}=4t^3$

The slope of the tangent line to the curve at $t=1$ is then

$\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2$

so the slope of the normal line is $-\frac12$ . When $t=1$ , we have

$x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2$

$y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2$

so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is

$y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3$

Q3. Differentiating with the product, power, and chain rules, we have

$y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4$

The derivative vanishes when

$\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23$

Q4. Differentiating with the product and chain rules, we have

$y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}$

The stationary points occur where the derivative is zero.

$-4xe^{-2x} = 0 \implies x = 0$

at which point we have

$y = (2x+1)e^{-2x} \bigg|_{x=0} = 1$

so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at $x=0$ .

$\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0$

The negative sign tells us this stationary point is a local maximum.

Q5. Differentiating the volume equation implicitly with respect to $t$ , we have

$V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$

When $r=5\,\rm cm$ , and given it changes at a rate $\frac{dr}{dt}=-1.5\frac{\rm cm}{\rm s}$ , we have

$\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}$

Q6. Given that $V=400\pi\,\rm cm^3$ is fixed, we have

$V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}$

Substitute this into the area equation to make it dependent only on $r$ .

$A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r$

Find the critical points of $A$ .

$\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}$

Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).

$\dfrac{d^2A}{dr^2}\bigg|_{r=2\sqrt[3]{50}} = \left(2\pi + \dfrac{1600\pi}{r^3}\right)\bigg|_{r=2\sqrt[3]{50}} = 6\pi > 0$

Hence the minimum surface area is

$A\bigg_{r=2\sqrt[3]{50}\,\rm cm} = \left(\pi r^2 + \dfrac{800\pi}r\right)\bigg|_{r=2\sqrt[3]{50}\,\rm cm} = 60\pi\sqrt[3]{20}\,\rm cm^2$

Q7. The volume of the box is

$V = 8x^2$

(note that the coefficient 8 is measured in cm) while its surface area is

$A = 2x^2 + 12x$

(there are two $x$ -by- $x$ faces and four 8-by- $x$ faces; again, the coefficient 12 has units of cm).

When $A = 210\,\rm cm^2$ , we have

$210 = 2x^2 + 12x \implies x^2 + 6x - 105 = 0 \implies x = -3 \pm\sqrt{114}$

This has to be a positive length, so we have $x=\sqrt{114}-3\,\rm cm$ .

Given that $\frac{dx}{dt}=0.05\frac{\rm cm}{\rm s}$ , differentiate the volume and surface area equations with respect to $t$ .

$\dfrac{dV}{dt} = (16\,\mathrm{cm})x \dfrac{dx}{dt} = (16\,\mathrm{cm})(\sqrt{114}-3\,\mathrm{cm})\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{4(\sqrt{114}-3)}5 \dfrac{\rm cm^3}{\rm s}$

$\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}$

5 0

1 year ago

Eddie is playing hide-and-seek with Harold and Gavin. Harold is hiding 12 meters south

natima [27]

Eddie and Gavin are 40 meters apart.

6 0

2 years ago

Solve for x.

Ugo [173]

Area of rectangle = L × B

Where

L is length and B is breadth

A = x × 5

45 = x × 5

x =45 /5

x = 9/1

x = 9 (units)

4 0

2 years ago

Read 2 more answers

Two mechanics Worked on a car The first mechanic charge $55 per hour The second mechanic charge $80 per hour the mechanics work

Tems11 [23]

a = hours worked by the first mechanic

b = hours worked by the second mechanic.

since the first mechanic charges $55 per hour, then for "a" hours that'd be a total of 55*a or 55a, likewise, for the second mechanic that'd be a total charge of 80*b or 80b.

we know all hours combined are 15, so then a + b = 15.

we also know that all charges combined are $950, so 55a + 80b = 950.

$\bf \begin{cases} a+b=15\\ \boxed{b}=15-a\\ \cline{1-1} 55a+80b=950 \end{cases}\qquad \qquad \stackrel{\textit{substituting on the 2nd equation}}{55a+80\left( \boxed{15-a} \right)=950} \\\\\\ 55a+1200-80a=950\implies -25a+1200=950\implies -25a=-250 \\\\\\ a=\cfrac{-250}{-25}\implies \blacktriangleright a = 10\blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{b=15-a\implies }b=15-10\implies \blacktriangleright b=5 \blacktriangleleft$

5 0

3 years ago