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2 years ago

Two angles are supplementary. The larger angle is 15 more than 10 times the smaller angle. FInd the measure of each angle.

1 answer:

7 0

$\text{Larger angle} =x\\\\\text{Smaller angle} =y\\ \\\text{According to the condition,}\\\\x=10y+15\\\\\text{The two angles are supplementary, so their sum is}~ 180^{\circ} \\\\x+y=180\\\\\implies 10y+15 +y = 180\\\\\implies 11y = 180-15\\\\\implies 11y=165\\\\\implies y = \dfrac{165}{11}\\\\\implies y = 15\\\\\text{So,} ~x =180-15 = 165\\\\\text{Hence the larger angle is}~ 165^{\circ}~ \text{and the smaller angle is }15^{\circ}$

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The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies No

Greeley [361]

Answer:

For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 0.02 g/mi

Standard Deviation, σ = 0.01 g/mi

Sample size, n = 81

We are given that the distribution of level of nitrogen oxides is a bell shaped distribution that is a normal distribution.

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.01}{\sqrt{81}} = 0.0011$

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.01

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 0.02}{0.0011})=0.01$

$= 1 -P( z \leq \displaystyle\frac{x - 0.02}{0.0011})=0.01$

$=P( z \leq \displaystyle\frac{x - 0.02}{0.0011})=0.99$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 0.02}{0.0011} = -2.326\\\\x = 0.0174$

For a level of 0.0174 or more of nitrogen oxide, the probability of fleet is 0.01.

5 0

2 years ago

Which is the graph of y = 2.5x? A. B. C. D.

Thepotemich [5.8K]

The answer is D. because first you got solve the problem.

y=2.5x
(turn y to 0)
0= 2.5x
÷2.5 ÷2.5
_________
0=x

So we got one answer (0,0)

y=2.5x
(turn x to 0)
y= 2.5(0)
y=2.5

(2.5, 0)

When you do (0,0) (2.5,0) on the graph it will look like D.)

5 0

3 years ago

This quiz is due in a hour I'll up the points to 25

Irina-Kira [14]

Answer:

The sin A expressed in ratio form is $\frac{\sqrt{6} }{4}$

Step-by-step explanation:

From the given right angled triangle diagram, sin A is calculated as follows;

$sin \ A = \frac{Opp}{hypo} \\\\sin \ A = \frac{2\sqrt{6} }{8} \\\\sin \ A = \frac{\sqrt{6} }{4}$

Therefore, the sin A expressed in ratio form is $\frac{\sqrt{6} }{4}$

3 0

3 years ago

Name the line and plane shown in the diagram. A.AB and plane ABD B.BA and plane DC C.AC and plane ABD D.AB and plane DA

elena-14-01-66 [18.8K]

Answer: The line is AB and the plane is ABD, the first option is the correct one.

Step-by-step explanation:

Ok, first some definitions.

A line is any line that crosses two colinear points. Particularly, you can see in the graph that the line crosses through A and B, so the line is AB.

A plane needs 3 non-colinear points (if the points where colinear, then the points may define a line). Other definition of plane is "a line and a point that is not in the line"

So, if our line is AB, then the possible planes are:

ABC and ABD.

then the correct option is:

Line AB and plane ABD, so the correct option is the first one.

6 0

2 years ago

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

6 0

2 years ago