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tatyana61 [14]
3 years ago
11

Two buses leave town at 1085 km apart at the same time and travel toward each other one bus travels 9 km an hour faster than the

y me in five hours what is the rate of each bus
Mathematics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

I am dababy ye ye clongne

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Over what interval is f(x) negative? Justify your answer.
tensa zangetsu [6.8K]

Answer in interval notation is  (-3, 1)

This is the same as saying -3 < x < 1

=====================================================

Explanation:

Since y = f(x), this means we're looking for points on the red curve such that they have a negative y coordinate. Visually, we're looking for points that are below the horizontal x axis.

These points are shown in the diagram below. I've marked the section we're after in blue. This blue section spans from x = -3 to x = 1. We exclude both endpoints because each x value leads to f(x) = 0, but we want f(x) values smaller than zero.

6 0
3 years ago
Please help with question :`D
olga55 [171]

Answer:

5.29 * 10^-5 is written as scientific notation.

For example:

9200 ---> 9.2 * 10³

123000 ---> 1.23 * 10^{5}

8640000 --->  8.64 * 10^6

where first number/letter should be between 1 - 10.

4 0
3 years ago
Which property is used in the problem below?
Natali [406]

Answer:

Distributive Property

Step-by-step explanation:

2(x+4) = 2x + 8

2x + 2(4) = 2x + 8

2x + 8 = 2x + 8

Henceforth, distributive property

7 0
2 years ago
What is the answer ?
labwork [276]

Answer:

arc VW = 66°

Step-by-step explanation:

The inscribed angle WXY is half the measure of its intercepted arc WY, so

WY = 2 × 57° = 114°

Arc YX = 180° ( semi- circle ) , then

VW = YX - WY = 180° - 114° = 66°

3 0
3 years ago
Read 2 more answers
Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8
Blababa [14]
w(s,t)=f(u(s,t),v(s,t))

From the given set of conditions, it's likely that you are asked to find the values of \dfrac{\partial w}{\partial s} and \dfrac{\partial w}{\partial t} at the point (s,t)=(1,0).

By the chain rule, the partial derivative with respect to s is

\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}

and so at the point (1,0), we have

\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial &#10;u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial &#10;v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21

Similarly, the partial derivative with respect to t would be found via

\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial &#10;u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial &#10;v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5
6 0
4 years ago
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