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Lapatulllka [165]
3 years ago
11

Help please today the teachers are grading it help!!!

Mathematics
2 answers:
Dima020 [189]3 years ago
8 0
6. 3.0 x 10^8

7. y = 3.5x + 7

8. 8 x 10^5 = 800000
    4 x 10^6 = 4000000
    
    4000000 / 800000 = 5 <==
Lemur [1.5K]3 years ago
3 0
#6
D. 3.0 x 10^8

#7
A. y = 3.5x + 7

#8
B. 5
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2.789 in word form and in expanded form
mylen [45]
Word form:
Two and seven hundred eighty-nine thousandths

Expanded form:

2

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The formula to determine energy is uppercase E = one-half m v squared. What is the formula solved for v?
alexandr402 [8]

Answer:

Step-by-step explanation:

E=\frac{1}{2}mv^2

All the variables on the right are being multiplied together then the whole mess is being divided by 2.  Let's get rid of the 2 first.  The undoing of division is multiplication, so we will begin by multiplying both sides by 2 to get

2E=mv^2

Next we will move the m. The undoing of multiplication is division.  So we divide both sides by m to get

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3 years ago
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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

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               =\frac{24\times (-13)}{26}

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Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

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2 years ago
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Mike had 25 stickers left

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