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3 years ago

6

Are there 4 feet in a yard ?

2 answers:

AysviL [449]3 years ago

5 0

No there are 3 feet in a yard

daser333 [38]3 years ago

5 0

Its not 4 feet it is actually 3 feet in a yard

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State the number of complex zeros, the possible number of real and imaginary zeros, and the possible rational zeros for each fun

ArbitrLikvidat [17]

Since the degree of this polynomial is 5, there will be 5 possible zeros. To find the possible rational 0s, use the rational root theorem (p/q). P is the last, non x value, which here it is the four on the end. The q is the leading coefficient, which is also q. Next, find all of the factors of q and p, which since they are both 4, are ±1, ±2, and ±4. Next do all possible values of p/q, which are ±1, ±2, ±4, ±1/2, and ±1/4. These are all your possible rational zeros. complex 0s only come in pairs, so the maximum there can be is 4 complex zeros, meaning there is at least one rational, real 0. (i graphed it it is -1/2, so all others must be rational or imaginary)

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4 years ago

Can someone help please?

deff fn [24]

Slope = 75
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4 years ago

Y = -6x + 1<br> Gradient (m) =<br> Y-intercept (c) =

Aloiza [94]

Answer:

answer is 7x add the this number

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3 years ago

Read 2 more answers

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

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

$=\frac{24\times (-13)}{\sqrt{676}}$

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

= -12 m/s

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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3 years ago

A store starts the day with 45 packages of cheese slices. Each package contains the same number of cheese slices. By the end of

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Answer

Step-by-step explanation:

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