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Alex17521 [72]
3 years ago
15

Los Angeles is about 385 miles from San Francisco. How far apart would the cities be on a map with a scale of 1 in. = 15 mi? If

necessary, round to the nearest hundredth.
A. 26.33 in.
B. 25.67 in.
C. 24.33 in.
D. 27.50 in.
Mathematics
2 answers:
beks73 [17]3 years ago
5 0
25.67 because 385/15 = 25.6666666666 and if you round you get answer B.

Lisa [10]3 years ago
4 0

Distance between Los Angeles and San Francisco = 385 miles

So , if we have to locate 385 miles on the map taking the scale

1 inches = 15 miles

1 miles = \frac{1 }{15} inches

so, 385 miles = \frac{1}{15}\times 385=25.666....

→→385 miles = 25.67 (approx)

Option (B) is correct.

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Solve the problem. The half-life of plutonium-234 is 9 hours. If 50 milligrams is present now, how much will be present in 6 day
MaRussiya [10]

Answer:

  4)  0.001

Step-by-step explanation:

The amount (a) present in h hours is ...

  a(h) = (initial value)×(1/2)^(h/(half-life))

6 days is 6×24 hours, so the amount remaining is ...

  a(6×24) = 50×(1/2)^(6×24/9) = 50/2^16 ≈ 0.001 . . . . milligrams

3 0
3 years ago
In the construction of the circumcenter, why do you only need to construct two perpendicular bisectors, and not all three?
Wittaler [7]
We know that the Circumcenter is the intersection of a triangle's right (perpendicular) bisectors. Once you have found two of these you can determine their point of intersection. 

It is a proven mathematical concept that the third perpendicular bisector will also cross through this intersection point. While it is not necessary to construct all three it is often done so to verify that the intersection of the first two perpendicular bisectors was in fact correct.
7 0
3 years ago
On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove
gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

v = \frac{d}{t}

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time t_1, distance of d_1, v = 40. So

v = \frac{d}{t}

40 = \frac{d_1}{t_1}

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is v = 60.

On the first day, he drove 2 more hours, which means that for the second day, the time is: t_1 - 2

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: d_1 - 20

Thus

v = \frac{d}{t}

60 = \frac{d_1 - 20}{t_1 - 2}

System of equations:

Now, from the two equations, a system of equations can be built. So

40 = \frac{d_1}{t_1}

60 = \frac{d_1 - 20}{t_1 - 2}

Find the total distance traveled in the two days:

We solve the system of equation for d_1, which gets the distance on the first day. The distance on the second day is d_2 = d_1 - 20, and the total distance is:

T = d_1 + d_2 = d_1 + d_1 - 20 = 2d_1 - 20

From the first equation:

d_1 = 40t_1

t_1 = \frac{d_1}{40}

Replacing in the second equation:

60 = \frac{d_1 - 20}{t_1 - 2}

d_1 - 20 = 60t_1 - 120

d_1 - 20 = 60\frac{d_1}{40} - 120

d_1 = \frac{3d_1}{2} - 100

d_1 - \frac{3d_1}{2} = -100

-\frac{d_1}{2} = -100

\frac{d_1}{2} = 100

d_1 = 200

Thus, the total distance is:

T = 2d_1 - 20 = 2(200) - 20 = 400 - 20 = 380

The total distance traveled in two days was of 380 miles.

For the relation between velocity, distance and time, you can take a look here: brainly.com/question/14307500

3 0
3 years ago
Need help solving algebraically!
alekssr [168]
We need to account for both x values on either side of the length, and width.
Thus, the length becomes 10 + x + x = 10 + 2x
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For the second question, I'm assuming we don't account for the area that is covered by the garden.
Then we can say that the path is measured by: (5 + 2x)(10 + 2x) - 50, which is the area of the garden itself.

(5 + 2x)(10 + 2x) - 50 = 54

Expanding the brackets:
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x(2x - 3) + 9(2x - 3) = 0

(x + 9)(2x - 3) = 0
x = -9, or x = 3/2

Since x > 0, then x ≠ -9
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6 0
3 years ago
Rewrite in simplest rational exponent form x^1/2*X^1/4
jasenka [17]

Answer:

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Step-by-step explanation:

First, let's examine our original statement.

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Using exponent rules, we know that if we have x^a \cdot x^b, then simplified, the answer will be equivalent to x^{a+b}.

So we can simplify this by adding the exponents \frac{1}{2} and \frac{1}{4}.

Converting  \frac{1}{2} into fourths gets us \frac{2}{4}.

\frac{2}{4} + \frac{1}{4} = \frac{3}{4}.

So we now have x^{\frac{3}{4}}.

When we have a number to a fraction power, it's the same thing as taking the denominator root of the base to the numerator power.

Basically, this becomes

\sqrt[4]{x^3}. (The numerator is what we raise x to the power of, the denominator is the root we take of that).

Hope this helped!

4 0
3 years ago
Read 2 more answers
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