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3 years ago

What is the value of t in the following equation?

1 answer:

3 0

Answer: t = 2698 as question is written. Just subtract 14 from both sides. There is either a typo in the question or the answer choices.

Step-by-step explanation:

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Slope for (-5,5) (5,-5)

kap26 [50]

Answer:

Slope(m)=-1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the distance between the points E(3, 7) and F(6, 5). The exact distance between the two points is .

S_A_V [24]

Answer:

$\sqrt{13}$

Step-by-step explanation:

The trick here is to know the distance formula (a formula that helps you find the distance between two points).

Distance formula: distance = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

But now that we know this, we can simply plug in our values and solve!

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\sqrt{(6-3)^2+(5-7)^2}\\\sqrt{(3)^2+(-2)^2}\\\sqrt{9+4}\\\sqrt{13}$

Now we can tell from our answer that the distance between both points is $\sqrt{13}$

7 0

3 years ago

PLEASE HELP SOON I DONT UNDERSTAND THIS QUIZ

Fantom [35]

1. h₀ = 70

v₀ = 0 since the rock is dropped it has no initial velocity

when the rock hits the ground the height = 0

plug into the formula

0 = -4.9t² + 0(t) + 70

0 = -4.9t² + 70

Add 4.9t² to both sides

4.9t² = 70

Divide both sides by 4.9

t² = 14.2857142857

Take the square root of both sides

t = 3.7796447301

t ≈ 3.8 seconds

I am not doing the rest I only did one just to help you understand a bit more

6 0

3 years ago

Abby caught a fish that was 10 3/8 inches long. Brandon caught a fish that was 8 5/8 inches long. How much longer was Abby's fis

Vilka [71]

what you do is take 5 away from 3. No! Borrow 1 Whole (9) and makes it 9 13/8 - 8 5/8 then subtract the 2 fractions

$\frac{13}{8} - \frac{5}{8} ...$

and that is

$\frac{8}{8}$

then convert that into 1 whole. Now add the whole to the 9 and thats 10. Now, subtract the 2 whole numbers. 10-8=2 so the answer is 2

5 0

3 years ago

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

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

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$ , the slope is again $-\frac{16}{64}=-\frac14$ , so the normal has slope +4, and so on.

5 0

2 years ago