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vfiekz [6]
1 year ago
15

HELP 100 POINTS WILL GIVE BRANLIEST IF ITS RIGHT LOOK IMAGE

Mathematics
1 answer:
____ [38]1 year ago
7 0

The value of m\angle2 is 132 and the value of y is 30.

19) In this question given that,

m\angle1=x+10

m\angle2=3x+18

We have to find the value of m\angle2.

As we know that the value of straight line is 180^o.

From the given figure we can see that

m\angle1+m\angle2=180^o

Now putting the value of m\angle1 and m\angle2 in the m\angle1+m\angle2=180^o.

x+10+3x+18=180^o

Now simplifying the expression.

4x+28=180^o

Subtract 28 on both side

4x+28-28=180^o-28

4x=152

Divide by 4 on both side

\frac{4}{4}x=\frac{152}{4}

x=38

We have to find the value of m\angle2.

As we know that m\angle2=3x+18.

Now putting the value of x in the given value of m\angle2.

m\angle2=3\times38+18

m\angle2=114+18\\

m\angle2=132

Hence, the value of m\angle2 is 132.

20) In this question we have to find the value of y.

As we know that the value of straight line is 180^o.

From the given figure we can see that

2y^o+(3y+30)^o=180^o

Now solving the expression to find the value of x.

2y^o+3y^o+30^o=180^o

Simplifying the expression.

5y^o+30^o=180^o

Subtract 30^o on both side

5y^o+30^o-30^o=180^o-30^o

Simplifying

5y^o=150^o

Divide by 5 on both side

\frac{5}{5}y^o=\frac{150^o}{5}

y^o=30^o

Hence, the value of y is 30.

To learn more about angle link is here

brainly.com/question/28451077

#SPJ1

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<u>Answer:</u>

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

<u>Solution:</u>

Given, Elena bikes 20 minutes each day for exercise.  

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,  

We know that, distance travelled = speed \times time

a. At a constant speed of 13 miles per hour for the entire 20 minutes  

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\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }

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\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

D = M mph \times 5 minutes + N mph \times 15 minutes

\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }

\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

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