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Hoochie [10]
2 years ago
4

Chris and Amy leave from work and drive in different directions. Their paths are at a 90-degree angle from one another, as shown

in the following illustration.
Chris is traveling at 40 mph. Amy leaves two hours later and is traveling at 60 mph. Assuming that t represents the time that Chris has been driving, which of the following equations can be used to calculate the values of t for which the distance between Chris and Amy is 300 miles?

A.) Square root (40t)^2-[60(t-2)]^2=300
B.) Square root (40t)^2+[60(t-2)]^2=300
C.) Square root (40t)^2+[60(t+2)]^2=300
D.) Square root (60t)^2+[40(t-2)]^2=300
Mathematics
1 answer:
AlladinOne [14]2 years ago
7 0
We know that

<span>the drawing is not necessary to solve the problem</span>

t-----------> <span>represents the time that Chris has been driving
</span>so
Chris distance is
y=40*t

Amy distance is
y=60*(t-2)

<span>applying the Pythagorean theorem
</span>let
a----------> Chris distance
b----------> Amy distance
c----------> <span>distance between Chris and Amy-------> 300 miles

c</span>²=a²+b²
so
300²=[40*t]²+[60*(t-2)]²--------> 300=√{[40*t]²+[60*(t-2)]}

the answer is the option
B.) Square root (40t)^2+[60(t-2)]^2=300

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grigory [225]

Answer:

5\ \frac{miles}{hour}

Step-by-step explanation:

<u><em>The complete question is</em></u>

The table shows how far a distance runner has traveled since the race began. What is her average speed, in miles per hour, during the interval 0.75 to 1.00 hours?

Time Elapsed (Hours) Miles Traveled (Miles)

0.50 2.00

0.75 3.50

1.00 4.75

Let

x ----> the time in hours

f(x) ----> the distance traveled in miles

so

we have

x (Hours)   0.50   0.75  1.00

f(x) (Miles)  2.00   3.50  4.75

we know that

To find the average speed, we divide the change in the output value by the change in the input value

the average speed is equal to

\frac{f(b)-f(a)}{b-a}

In this problem we have

a=0.75

b=1.00

f(a)=f(0.75)=3.50  

f(b)=f(1.00)=4.75

Substitute

\frac{4.75-3.50}{1-0.75}

\frac{1.25}{0.25}

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8 0
2 years ago
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Solve for q. 22q+73 &gt; 52q + 63
sammy [17]
Hey!

Let's write the problem.
22q+73\ \textgreater \ 52q+63
Subtract 73 from both sides.
22q+73-73\ \textgreater \ 52q+63-73
22q\ \textgreater \ 52q-10
Subtract 52q from both sides.
22q-52q\ \textgreater \ 52q-10-52q
-30q\ \textgreater \ -10
Since we don't want negative, we will multiply both sides by -1.
\left(-30q\right)\left(-1\right)\ \textless \ \left(-10\right)\left(-1\right)
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Divide both sides by 30.
\frac{30q}{30}\ \textless \ \frac{10}{30}

Our final answer would be,
q\ \textless \ \frac{1}{3}

Thanks!
-TetraFish 
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Vikki [24]

Answer:

4 balloons for each balloon animal.

Step-by-step explanation:

Amira sells balloon animals. The given table compares the number of balloon animals sold and the remaining number of balloons on a certain day.

Animals        Balloons

15                   200

24                   164

33                   128

When she sold 15 animals the leftover balloons were 200.

then she sold 24 animals the number of leftover balloons were 164.

The difference of the number of items = 24 - 15 = 9 more animals sold

The difference of the leftover balloons = 200 - 164 = 36 balloons used

Now the number of her sold animals was = 33

and the left over balloons = 128

The number of sale was increased by 24 to 33 = 33 - 24 = 9 more animals

and the leftover balloons now 164 to 128 = 164 - 128 = 36 used

Therefore, we can see each sale of 9 balloon animals she used 36 balloons.

so Amira used for each balloon animal = \frac{36}{9}

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Amira used 4 balloons for each balloon animal.

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torisob [31]
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Step-by-step explanation:

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