Eight less than a number y is ninety four

Can someone help me immediately on this question having to do with triangle ABC?

<h3>Answers:</h3><h3>a. Vertices of triangle ABC are: A, B, C</h3><h3>b. Sides of triangle ABC are: AB, BC, AC</h3><h3>c. The side between angle A and angle C is: side AC</h3><h3>d. The angle between sides AB and CA is: angle A</h3><h3>e. Scalene triangle</h3>

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Explanations:

a. Each uppercase letter represents a point or angle of the triangle.
b. Connect two points of a triangle and you get a line segment. The order of the letters does not matter. So AB is the same as BA.
c. Like with part b, connecting two angles or points forms a segment.
d. Note how the letter "A" is in both AB and CA, so this is the shared angle between the two segments.
e. Sides AB, BC, and AC are all different lengths, so we have a scalene triangle. If you had two sides equal to each other, then you'd have an isosceles triangle. If all three sides are equal, then it would be equilateral.

There is no need for a diagram, but if you want, you can draw one out. See the attached image below for the diagram. This diagram should hopefully answer any questions you may have about the explanations above. There are many ways to draw the triangle, so your diagram might look different from mine.

7 0

2 years ago

A batch of 40 semiconductor chips is inspected by choosing a sample of 5 chips. Assume 10 of the chips do not conform to custome

butalik [34]

Answer:

a) 658008 samples

b) 274050 samples

c) 515502 samples

Step-by-step explanation:

a) How many ways sample of 5 each can be selected from 40 is just a combination problem since the order of selection isn't important.

So, the number of samples = ⁴⁰C₅ = 658008 samples

b) How many samples of 5 contain exactly one nonconforming chip?

There are 10 nonconforming chips in the batch, and 1 nonconforming chip for the sample of 5 be picked from ten in the following number of ways

¹⁰C₁ = 10 ways

then the remaining 4 conforming chips in a sample of 5 can be picked from the remaining 30 total conforming chips in the following number of ways

³⁰C₄ = 27405 ways

So, total number of samples containing exactly 1 nonconforming chip in a sample of 5 = 10 × 27405 = 274050 samples

c) How many samples of 5 contain at least one nonconforming chip?

The number of samples of 5 that contain at least one nonconforming chip = (Total number of samples) - (Number of samples with no nonconforming chip in them)

Number of samples with no nonconforming chip in them = ³⁰C₅ = 142506 samples

Total number of samples = 658008

The number of samples of 5 that contain at least one nonconforming chip = 658008 - 142506 = 515502 samples

6 0

2 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

Stanley is having wood flooring installed in his living room. The installer has used boxes of flooring on of the room. How many

Ray Of Light [21]

<h3>Total approximately 16 boxes of flooring will be needed in Stanley's living room. </h3>

Step-by-step explanation:

Here the given question is <u>incomplete</u>.

Stanley is having wood flooring installed in his living room. The installer has used 10 1/2 boxes of flooring on 2/3 of the room. How many total boxes of flooring will be needed for Stanley's living room?

The number of boxes used in the two third of room = $10 (\frac{1}{2} ) = 10.5$ boxes

So, $(\frac{2}{3} )$ room = 10. 5 boxes

So, 1 room = $10.5 \times (\frac{3}{2} ) = 15.75 \approx 16$

So, total approximately 16 boxes of flooring will be needed in Stanley's living room.

6 0

3 years ago

.

tankabanditka [31]

The function is continuous

7 0

3 years ago