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Vlada [557]
3 years ago
9

H=vt-5t^2, solve for v

Mathematics
2 answers:
Kamila [148]3 years ago
7 0
H=vt-5t^2 for v
h=t(v-5t)
h/t=v-5t
h/t+5t=v
(h+5t^2)/(t)=v

Answer: v=(h+5t^2)/(t)
Colt1911 [192]3 years ago
4 0

Answer:

The value of v is v=\frac{h+5t^2}{t}    

Step-by-step explanation:

Given : Expression h=vt-5t^2  

To find : Solve for v?

Solution :

Step 1 - Write the expression,

 h=vt-5t^2

Step 2 - Add 5t^2 both side,

 h+5t^2=vt-5t^2+5t^2

 h+5t^2=vt

Step 3 - Divide by t both side,

 \frac{h+5t^2}{t}=\frac{vt}{t}

 v=\frac{h+5t^2}{t}

Therefore, The value of v is v=\frac{h+5t^2}{t}

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Answer:

just look at the decimal after the decimal number is the tenths place the hundreds place and so on if i have to round 567.1092 i would round the nine up to make the zero a one so my rounded answer would be 567.11

Step-by-step explanation:

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3 years ago
Please help me with these calculus bc questions
zhannawk [14.2K]

4. Compute the derivative.

y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1

Find when the gradient is 7.

4x - 1 = 7 \implies 4x = 8 \implies x = 2

Evaluate y at this point.

y = 2\cdot2^2-2-1 = 5

The point we want is then (2, 5).

5. The curve crosses the x-axis when y=0. We have

y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4

Compute the derivative.

y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}

At the point we want, the gradient is

\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}

6. The curve crosses the y-axis when x=0. Compute the derivative.

\dfrac{dy}{dx} = 3x^2 - 4x + 5

When x=0, the gradient is

\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}

7. Set y=5 and solve for x. The curve and line meet when

5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92

Compute the derivative (for the curve) and evaluate it at these x values.

\dfrac{dy}{dx} = 4x + 7

\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}

\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}

8. Compute the derivative.

y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b

The gradient is 8 when x=2, so

2a\cdot2 + b = 8 \implies 4a + b = 8

and the gradient is -10 when x=-1, so

2a\cdot(-1) + b = -10 \implies -2a + b = -10

Solve for a and b. Eliminating b, we have

(4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}

so that

4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}.

5 0
2 years ago
Tulsi and Gary are competing in a basketball free throw competition. Tulsi’s ratio of free throws made to free throws missed is
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Gary because 10 to 15 rounds down to 2/3 which is greater than 3 to 5
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Read 2 more answers
8 * 5/7 equalfind the product write the product in its simplest form 8 * 5/7 ​
Naily [24]

The product of 8 and 5/7 is 5.7

<u>Step-by-step explanation:</u>

The given product is 8 \times 5/7.

<u>In the product, two terms are given :</u>

  • The first term 8 is a whole number.
  • The second term 5/7 is a fraction.

1) So, to find the product of the first term and the second term, one method is to simplify the fraction into decimal number and then perform multiplication operation.

2) Either you can multiply the two terms in the first step and then the division operation is performed at the final step.

<u>First method :</u>

<u><em>Step 1 :</em></u> Convert the fraction into decimal

⇒ 5/7 = 0.71

<u><em>Step 2 :</em></u> Multiply the result with the other term.

⇒ 8 × 0.71

⇒ 5.68 (approximately 5.7)

<u>Second method :</u>

<u><em>Step 1 :</em></u> Multiply both the terms

⇒ 8 ×(5/7) = 40/7

<u><em>Step 2 :</em></u> divide 40 by 7

⇒ 40÷ 7 = 5.7

Therefore, the product of 8 and 5/7 is 5.7

7 0
4 years ago
What is the slope of the line?
crimeas [40]
You’re close already :).
Slope is rise/run!
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