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Mekhanik [1.2K]
3 years ago
7

By the time shana had completed 3/8 of her first lap in a race, she had also completed 1/32 of the entire race. How many laps we

re there in the race.
Mathematics
1 answer:
Yuri [45]3 years ago
3 0
3/8 of a lap is 1/32 of the entire race.
x - the number of laps in the race

\frac{3}{8}=\frac{1}{32}x \\
\frac{3}{8} \times 32=x \\
3 \times 4=x \\
x=12

There were 12 laps in the race.
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Solve for x. round to the nearest hundredth if necessary.
iris [78.8K]

Answer: x=11.47

Step-by-step explanation:

Given the angle of 55 degrees, you know that the adjacent side is "x" and the length of the hypotenuse is 20.

Therefore, you need to remember the following identity:

cos\alpha=\frac{adjacent}{hypotenuse}

Then, knowing that:

\alpha=55\°\\adjacent=x\\hypotenuse=20

 You need to substitute these values intocos\alpha=\frac{adjacent}{hypotenuse}:

 cos(55\°)=\frac{x}{20}

Now, you can solve for "x":

20*cos(55\°)=x\\x=11.471

Rounded to the nearest hundreth:

x=11.47

6 0
3 years ago
4. Two points on a line are (-10, 1) and (5, -5). If
Vsevolod [243]

Answer:

The x-coordinate of another point is zero

Step-by-step explanation:

step 1

Find the slope between the two given points

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

we have

(-10,1),(5,-5)

substitute in the formula

m=\frac{-5-1}{5+10}

m=\frac{-6}{15}

Simplify

m=-\frac{2}{5}

step 2

Find the x-coordinate of another point

we have

(x,-3)

we know that

If the other point is on the line, then the slope between the other point and any of the other two points must be the same

so

Find the slope between the points

(x,-3),(5,-5)

Remember that

m=-\frac{2}{5}

substitute in the formula

-\frac{2}{5}=\frac{-5+3}{5-x}

-\frac{2}{5}=-\frac{2}{5-x}

the denominators must be the same

5=5-x

x=0

therefore

The x-coordinate of another point is zero

3 0
3 years ago
Find the approximate surface-area-to-volume ratio of a bowling ball with a radius of 5 inches.
Nataly_w [17]
Surface area of a ball: 
S=4πr^2=4*π*5^2=100π

Volume of a circle:
V=4/3*π*r^3=4/3*π*125=(500/3)*π=165π

The approximate surface-area-to-volume ratio would be 1:1,5
6 0
3 years ago
Trigonometry:
Trava [24]

A. The approximate height of the building is 100 m

B. We can use the arc length formula to obtain an approximation of BC since angle A is small.

A.

The approximate height of the building is 100 m

To find the approximate height of the building, we consider the diagram.

From the diagram, we have that using trigonometry,

tanA = BC/AB

Now

  • A = 0.05 radians,
  • BC = height of building and
  • AB = 2 km = 2000 m
<h3 /><h3>Finding the value of BC</h3>

So, making BC subject of the formua, we have

BC = ABtanA

Substituting the values of the variables into the equation, we have

BC = ABtanA

BC =2000tan0.05

BC = 2000 × 0.05

BC = 100 m

So, the approximate height of the building is 100 m

B.

We can use the arc length formula to obtain an approximation of BC since angle A is small.

<h3 /><h3>Arc Length Formula</h3>

We know that the arc length formula L = rФ where

  • r = radius and
  • Ф = angle in radians
<h3 /><h3>The approximate height</h3>

Now BC = ABtanA

We know that Ф ≅ tanФ when Ф is small.

<h3 /><h3>The comparison</h3>

So, BC = AB × A which is the arc length formula with

  • L = BC,
  • r = AB and
  • Ф = A

So, we can use the arc length formula to obtain an approximation of BC since angle A is small.

Learn more about approximate height of building here:

brainly.com/question/3144976

7 0
2 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
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