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If 
AD and 
PM are the medians of triangles 
/_\ABC and 
/_\PQR, respectively where 
/_\ABC∼/_\PQR, prove that 
(AB)/(PQ)=(AD)/(PM).
10. Find the unknown entries 
a,b,c and 
d in the following distribution of heights of students in the class.




Height (in cm)
f
cf



150-155
a
12



155-160
13
b



160-165
10
35



165-170
c
43



170-175
7
d





{:[d=50],[a=12],[b=25],[c=8]:}

Solve the following system of equations by elimination method:


0.2 x+0.3 y=1.3" and "0.4 x+0.5 y=2.3
Page 2 of Set 2

or If ADAD and PMPM are the medians of triangles ABC\triangle ABC and PQR\triangle PQR, respectively where ABCPQR\triangle ABC \sim \triangle PQR, prove that ABPQ=ADPM\frac{AB}{PQ} = \frac{AD}{PM}. 1010. Find the unknown entries a,b,ca, b, c and dd in the following distribution of heights of students in the class. Height (in cm) ff cfcf 150150155-155 PMPM00 1212 155155160-160 1313 PMPM11 160160165-165 1010 3535 165165170-170 PMPM22 4343 170170175-175 77 dd PMPM44 Solve the following system of equations by elimination method: PMPM55 and PMPM66 Page 22 of Set 22

Full solution

Q. or If ADAD and PMPM are the medians of triangles ABC\triangle ABC and PQR\triangle PQR, respectively where ABCPQR\triangle ABC \sim \triangle PQR, prove that ABPQ=ADPM\frac{AB}{PQ} = \frac{AD}{PM}. 1010. Find the unknown entries a,b,ca, b, c and dd in the following distribution of heights of students in the class. Height (in cm) ff cfcf 150150155-155 PMPM00 1212 155155160-160 1313 PMPM11 160160165-165 1010 3535 165165170-170 PMPM22 4343 170170175-175 77 dd PMPM44 Solve the following system of equations by elimination method: PMPM55 and PMPM66 Page 22 of Set 22
  1. Given Triangles ABCABC and PQRPQR: Given: Triangles ABCABC and PQRPQR are similar, so their corresponding sides are proportional.
  2. Proportional Sides: Since ABC ∼ PQR, we have:\newlineABPQ=BCQR=CARP \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}
  3. Medians Proportional: Medians of similar triangles are also proportional to the corresponding sides. So:\newlineADPM=ABPQ \frac{AD}{PM} = \frac{AB}{PQ}
  4. Height Interval 150150155-155 cm: Therefore, we have:\newlineABPQ=ADPM \frac{AB}{PQ} = \frac{AD}{PM}
  5. Height Interval 155155160-160 cm: Given cumulative frequency (cf) for the height interval 150150155-155 cm is 1212. So, a=12a = 12.
  6. Align Coefficients: Given cumulative frequency (cf) for the height interval 155155160-160 cm is b. Since the cumulative frequency for the next interval (160160165-165 cm) is 3535, we have:\newlineb=3513=22 b = 35 - 13 = 22
  7. Subtract Equations: Multiply the first equation by 22 to align coefficients:\newline2(0.2x+0.3y)=2(1.3) 2(0.2x + 0.3y) = 2(1.3) \newline0.4x+0.6y=2.6 0.4x + 0.6y = 2.6
  8. Solve for y: Subtract the second equation from the modified first equation:\newline(0.4x+0.6y)(0.4x+0.5y)=2.62.3 (0.4x + 0.6y) - (0.4x + 0.5y) = 2.6 - 2.3 \newline0.1y=0.3 0.1y = 0.3
  9. Substitute y into Equation: Solve for y:\newliney=0.30.1=3 y = \frac{0.3}{0.1} = 3
  10. Substitute y into Equation: Solve for y:\newliney=0.30.1=3 y = \frac{0.3}{0.1} = 3 Substitute y = 33 into the first equation:\newline0.2x+0.3(3)=1.3 0.2x + 0.3(3) = 1.3 \newline0.2x+0.9=1.3 0.2x + 0.9 = 1.3 \newline0.2x=0.4 0.2x = 0.4 \newlinex=0.40.2=2 x = \frac{0.4}{0.2} = 2

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