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iogann1982 [59]

3 years ago

14

Asia new boat has a cruising speed of 24 miles per hour at that rate how long will it take her to travel the 6 miles across old

hickory lake

1 answer:

Igoryamba3 years ago

3 0

Answer:

1/4 of an hour, or 60/4=15, 15 minutes.

Step-by-step explanation:

24/1=6/x

cross product

1*6=24*x

6=24x

x=6/24

x=1/4

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Alma is changing the shape of her backyard from 150 feet long by 62 feet wide to a square that has the same area. What is the pe

umka2103 [35]

Hey there!
Here's your answer: the perimeter of the redone yard would be ~193 feet

Hope this helps!

6 0

2 years ago

I need help and I’m too lazy to type this out tell me you can’t see the picture

wlad13 [49]

It’s really hard to see the picture

5 0

3 years ago

Evaluate the function fx = 4 * 7^t for x= -1 and x=2

Alex73 [517]

Well what i think you're asking here...I apologie if I'm wrong but what I did was

f(-1) = 4 * 7^t = 4/7

and

f(2) = 4 * 7^t = 196.

Is this what you wanted if not I can...try another way.

6 0

2 years ago

PolarNik [594]

Question 1:

<span>5z – 9 = 36

5z = 45

z = </span> $\frac{45}{5}$

z = 9

Hence, the answer is D.

Question 2:

<span>5(2z + 4) = 0
</span><span>
10z + 20 = 0

10z = -20

z = </span> $\frac{-20}{10}$
<span>
z = -2

Hence, the answer is A.

Question 3:

4a + 3(a + 1) = 24

4a + 3a + 3 = 24

7a = 21

a = </span> $\frac{21}{7}$

a = 3

Hence, the answer is B.

Question 4:

<span>3(4x + 5) = 12

12x + 15 = 12

12x = 12 - 15

12x = -3

x = </span> $\frac{-3}{12} = - \frac{1}{4}$

Hence, the answer is A.

Question 5:

<span>4x – 8 – 4x = 9

-8 </span>≠ 9

Hence, the answer is A.

8 0

3 years ago

Read 2 more answers

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

5 0

3 years ago