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3 years ago

Find the values of x and y.

Mathematics

1 answer:

Doss [256]3 years ago

6 0

Answer: y=37.5degrees and x=110degrees

Step-by-step explanation: sum of angle in a quadrilateral =360 and opposite angles=180. 115+2y=180,2y=180-115=75. 2y/2=75/2. therefore y=37.5. while x=70+x=180,x=180-70=110. x=110

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A ) 4.05 per pound <br> B) $5.40 per pound <br> C) $2.70 per pound <br> D) $1.35 per pound

Nadusha1986 [10]

Answer:

D is the correct answer

Step-by-step explanation:

As you see, the product of peaches 1 and 2 have the different in cost is $1.35 and $2.70. As well as the following product of peaches also have the same increase in money.

So 2.70 - 1.35 = 1.35 so the common increase would be $1.35

Hope this help you :3

6 0

3 years ago

Consider the initial value problem yâ€²+2y=4t,y(0)=8.

Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

$y'+2y=4t\\\\y(0)=8$

a) The algebraic equation obtain by using the Laplace transform is:

$L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\$

next, you replace (1) and (2):

$sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}$ (this is the algebraic equation)

$sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}$ (this is the solution for Y(s))

$y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}$

To find the inverse Laplace transform of the first term you use partial fractions:

$\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}$

Thus, you have:

$y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}$

(this is the solution to the differential equation)

5 0

3 years ago

What are the domain and range of f(x) = 2|x - 4|?

exis [7]

Answer:

D : All real numbers

R : (-∞,∞)

Step-by-step explanation:

When your x is in a set of absolute value sign, any number can work for x.

As with Domain, you can do the same with range. In this case, you can go from both positive infinity and negative infinity and you will always get a real number solution.

6 0

3 years ago

A car traveling at 30 miles per hour requires 90 feet to come to a cineplete stop d=distance s=speed and m is the constant

NeTakaya

just plug in the information:

d=90

s=30

so 90=m*30^2

90=m*900

now divide both sides by 900 to get m

m=0.1

6 0

3 years ago

67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

$P(X=r) = nCr p^r q^{n-r}$

where $nCr = \frac{n!}{(n-r)! r!}$

a) Probability that exactly 29 of them are spayed or neutered

$P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074$

b) Probability that at most 33 of them are spayed or neutered

$P(X \leq 33) =1 - P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66$

c) Probability that at least 30 of them are spayed or neutered

$P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79$

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

$P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574$

8 0

3 years ago