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bezimeni [28]
3 years ago
12

The inner square has sides of 3 inches and the outer square has sides

Mathematics
1 answer:
EastWind [94]3 years ago
7 0
X is the hypotenuse of an isosceles triangle. The length of the sides are the difference of the inner and outer squares divided by 2. Which means that the sides of the triangle is 1.5 inches. Using that information, the length of X can be deduced to be 2.12 inches using Pythagoras theorem
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(1+5i) + (-8+i) ? <br> Need answer and steps
Shtirlitz [24]

Answer:

6\sqrt{-1}-7

Step-by-step explanation:

simplyifying this we get

1+5i-8+i

= 6i-7

I am assuming the i you are talking about here is the sqrt of negative one so that would be 6\sqrt{-1}-7

7 0
3 years ago
The coordinates of the endpoint of QS are Q(-9,8) and S(9,-4). Point R is on cue as such that QR:RS Is in the ratio 1:2. What ar
marishachu [46]

R(–3, 4)

Step-by-step explanation:

Let Q(-9,8) and S(9,-4) be the given points and let R(x, y) divides QS in the ratio 1:2.

By section formula,

R(x, y)=R\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)

Here, x_{1}=-9, y_{1}=8, \text { and } x_{2}=9, y_{2}=-4 \text { and } m=1, n=2

Substituting this in the section formula

R(x, y)=R\left(\frac{1(9)+2(-9)}{1+2}, \frac{1(-4)+2(8)}{1+2}\right)  

To simplifying the expression, we get

\Rightarrow R(x, y)=R\left(\frac{9-18}{3}, \frac{-4+16}{3}\right)

\Rightarrow R(x, y)=R\left(\frac{-9}{3}, \frac{12}{3}\right)

⇒ R(x,y) = R(–3,4)  

Hence, the coordinates of point R is (–3, 4).

6 0
3 years ago
What is the equivalent to 3/20 as a fraction
aalyn [17]
6/40 is equalvilant because all you have to do is double or simplify it by 2
5 0
2 years ago
Read 2 more answers
O, R and S are points in the same horizontal plane. /OR/ = 20m and /OS/= 32m.The bearing of R and S from O are 030° and 135° res
Viktor [21]
<h3>Answer:  roughly 12.6274 meters</h3>

The more accurate value is 12.6274169979696, though it's not fully exact.

Round this however you need to.

The exact distance 32*sin(45) - 10 meters.

===========================================================

Explanation:

Refer to the diagram below.

I'll use point A in place of point O since the letter 'oh' is very similar looking to the number zero.

Plot point A at the origin (0,0). While at point A, look directly north. Then turn 30 degrees eastward to look at the bearing 030°. Next, move 20 meters along that bearing direction to arrive at point R. Segment AR is 20 meters long.

In the diagram, note how angle RAB is 30 degrees. The side opposite this is BR = m.

We can use the sine ratio to say that

sin(angle) = opposite/hypotenuse

sin(A) = BR/AR

sin(30) = m/20

m = 20*sin(30)

m = 10

-----------------------------------

While still at point A, look directly north and turn 135 degrees clockwise (ie toward the east) and move 32 meters along that bearing. You'll arrive at point S as the diagram shows.

Notice how

angle RAB + angle RAC + angle CAS = 30+60+45 = 135

The remaining angle DAS is 180-135 = 45 degrees.

When focusing on triangle DAS, we can say

sin(A) = DS/AS

sin(45) = n/32

n = 32*sin(45)

n = 22.6274169979696

This value is approximate.

----------------------------------

Subtract the values of m and n

n - m = 32*sin(45) - 10 = exact distance

n - m = 22.6274169979696 - 10

n - m = 12.6274169979696

n - m = 12.6274 = approximate distance

Round it however you need to. I'm choosing to round to four decimal places.

So we see that point S is roughly 12.6274 meters east of point R.

If your teacher wants the exact distance, then stick with 32*sin(45)-10.

8 0
3 years ago
The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. (a) Determine the
maxonik [38]

Answer:

a) 0.5 = 50% of flanges exceed 1 millimeter.

b) A thickness of 0.96 millimeters is exceeded by 90% of the flanges

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value higher than x is given by:

P(X > x) = \frac{b - x}{b - a}

The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.

This means that a = 0.95, b = 1.05

(a) Determine the proportion of flanges that exceeds 1.00 millimeters.

P(X > 1) = \frac{1.05 - 1}{1.05 - 0.95} = \frac{0.05}{0.1} = 0.5

0.5 = 50% of flanges exceed 1 millimeter.

(b) What thickness is exceeded by 90% of the flanges?

This is x for which:

P(X > x) = 0.9

So

\frac{1.05 - x}{1.05 - 0.95} = 0.9

1.05 - x = 0.9*0.1

x = 1.05 - 0.9*0.1

x = 0.96

A thickness of 0.96 millimeters is exceeded by 90% of the flanges

4 0
3 years ago
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