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3 years ago

What is the average speed if Time = 3 Distance = 90

Mathematics

1 answer:

Natasha2012 [34]3 years ago

6 0

Speed= distance divided by time. The answer is 30, and im assuming miles per hour. Hope this helps

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Ann, Chang, and Kareem sent a total of

Gnoma [55]

It is given in the question that

Ann, Chang, and Kareem sent a total of 89 text messages over their cell phones during the weekend. Ann sent 9 fewer messages than Chang. Kareem sent 3 times as many messages as Ann.

Let number of messages Chang sent be x. And Ann sent 9 fewer than Chang, so Ann sent (x-9) messages. And Kareem sent 3 times as many messages as Ann, so Kareem sent 3(x-9) messages.

And the total messages are 89. That is

$x+x-9+3(x-9)=89$

Combining like terms

$x + x - 9 + 3x-27 =89
\\
5x -36=89
\\
5x = 125
\\
x =25$

So Chang sent 25 messages, Ann sent

$25-9=16 messages$

And Kareem sent

$3*16 = 48 messages$

8 0

3 years ago

Can someone help me with question 9

Troyanec [42]

Please refer the pictures below-

Hence the area of the road is 9600m2

8 0

3 years ago

Find the quotient of 812.30 divided by 83. Round to the nearest tenth.

elena-14-01-66 [18.8K]

812.30/83 = 9.78674698.....
Nearest tenth means one number under decimal place so...

812.30/83 ~ 9.8

Explanation: The 9.7... rounded up to 9.8 because the number after the tenths place (in this case the number after the 7) is greater or equal to 5.

7 0

3 years ago

We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure

STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be $h(t) = -16\cdot t^{2} + 128\cdot t + 320$ , the first and second derivatives are, respectively:

First Derivative

$h'(t) = -32\cdot t +128$

Second Derivative

$h''(t) = -32$

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

$-32\cdot t +128 = 0$

$t = \frac{128}{32}\,s$

$t = 4\,s$ (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

$h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320$

$h(4\,s) = 576\,ft$

The highest altitude that the object reaches is 576 feet.

6 0

4 years ago

30 POINTS!! ASAP! PLS SHOW WORK

sp2606 [1]

Answer:

QII

Step-by-step explanation:

We can use the acronym: ASTC or All Students Take Calculus.

The first letter of each word tells us details within a certain quadrant.

All trig functions in QI will be positive.

Only sine (and cosecant) will be positive in QII.

Only tangent (and cotangent) will be positive in QIII.

And only cosine (or secant) will be positive in QIV.

We know that tan(θ)<0. In other words, it's negative.

So, it our angle θ <em>cannot</em> be in QI or QIII.

We also know that cos(θ)<0.

So, this removes QIV, since in QIV, cosine is positive.

Therefore, the only choices that remains is QII.

In QII, sine is positive, tangent is negative, and cosine is negative.

This fits all our conditions, so θ is in QII.

And we're done!

3 0

3 years ago