The two triangles are similar. What is the value for the ? ? Round to the nearest tenth if necessary.
L 28 35° K J ? R 42 35° P 33
Answers
Answer:
KJ = 22
Step-by-step explanation:
From the given diagram, since the figure are similar we can compare their sides.
So that,
[tex]\frac{JL}{PR}[/tex] = [tex]\frac{JK}{PQ}[/tex]
Let KJ be represented by x. But;
JL = 28, PR = 42 and PQ = 33
Thus,
[tex]\frac{28}{42}[/tex] = [tex]\frac{x}{33}[/tex]
x = [tex]\frac{28*33}{42}[/tex]
= [tex]\frac{924}{42}[/tex]
= 22
x = 22
The n,
KJ = 22
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Step-by-step explanation:
Given
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Answer:
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a. Find the length of the midsegment of an equilateral triangle with side lengths of 12.5 cm.
b. Given that UT is the perpendicular bisector of line segment AB, where T is on AB, find the length of AT given AT = 3x + 6 and TB = 42 - x.
c. Given angle EFG has angle bisector FH, where EF = GF, find the value of y if EH = 5y + 10 and HG = 28 - y.
Answers
Answer:
a.the length of midsegment of an equilateral triangle with side lengths of 12.5 cm.=12.5/2=6.25cm
b.
sinceUT is the perpendicular bisector of line segment AB, where T is on AB, find the length of AT given AT = 3x + 6 and TB = 42 - x.
than AT=BT
3x+6=42-x
3x+x=42-6
x=36/4=9
AT=3×9+6=27+6=33
c.
since angle EFG has angle bisector FH, where EF = GF, find the value of y if EH = 5y + 10 and HG = 28 - y.
EH=GH
5y+10=28-y
6y=28-10
y=18/6=3
a. The length of the midsegment of an equilateral triangle is 6.25 cm
b. The length of AT is 33 units.
c. The value of y is 3.
What is an equilateral triangle?
"It is a triangle whose all sides are equal in length and all angles measure 60° "
What is perpendicular bisector?"It is a line that intersects another line perpendicularly and divides it into wo equal parts."
What is angle bisector?"It is a line that splits an angle into two equal angle."
What is an isosceles triangle?"It is a triangle that has two sides of equal length."
What is midsegment of the triangle?"It is the line segment connecting the midpoints of the two sides of a triangle."
What is midsegment formula?"midsegment = 0.5 × base of triangle"
a.
We have been given an equilateral triangle with side lengths of 12.5 cm.
We need to find the length of the midsegment of an equilateral triangle.
Using midsegment formula,
⇒ midsegment = 0.5 × 12.5
⇒ midsegment = 6.25 cm
Therefore, the length of the midsegment of an equilateral triangle is 6.25 cm
b.
UT is the perpendicular bisector of line segment AB, where T is on AB
We know that the perpendicular bisector divides the line segment into two equal parts.
⇒ AT = TB
⇒ 3x + 6 = 42 - x
⇒ 3x + x = 42 - 6
⇒ 4x = 36
⇒ x = 9
So, the length of AT would be,
⇒ AT = 3x + 6
⇒ AT = 3(9) + 6
⇒ AT = 33 units
Therefore, the length of AT is 33 units.
c.
An angle EFG has angle bisector FH, where EF = GF
From given data, we can say that for it must an isosceles triangle EFG with EF = GF and base is EG.
An angle EFG has angle bisector FH and point H is on the side EG.
We know that the angle bisector of isosceles triangle is also the perpendicular bisector of the base of an isosceles triangle.
⇒ EH = HG
⇒ 5y + 10 = 28 - y
⇒ 5y + y = 28 - 10
⇒ 6y = 18
⇒ y = 3
Therefore, the value of y is 3.
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An apple has a mass of 32.06 grams. What is the mass of the apple rounded to the nearest tenth of a gram?
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Hope this helps! Have a nice life!The model represents 3 ÷ 1/5
Which multiplication can you use to check the answer?
Three squares are shown. Each square is divided into 5 fifths.
A. 3 x 1/5= 3/5
B. 15 x 1/5 = 3
C. 15 x 3/5 = 9
D. 15 x 5/3 =25
Answers
Answer:
the answer is 2
Step-by-step explanation:
The equation which represent the three squares which is divided into the 5 fifths is,
[tex]15 \times \dfrac{1}{5} = 3[/tex]
What is a multiple of fraction?A fraction number is the number which is a part of a whole number. It is written with a numerator and denominator. Fraction number are written as,
[tex]\dfrac{a}{b}[/tex]
Here (a) is the numerator and (b) is the denominator.
A multiple of a fraction is the number which is n times of the original function.
The model represents,
[tex]3 \div 1/5[/tex]
Here, the total number of square are given is 3. These three square are then divided into the one fifth.
Thus, we get total 3 times of 5 one fifth which is equal to the entire 3 squares
.
Thus, the equation can be written for this condition as,
[tex]15 \times \dfrac{1}{5} = 3[/tex]
Hence, the equation which represent the three squares which is divided into the 5 fifths is,
[tex]15 \times \dfrac{1}{5} = 3[/tex]
Option B is the correct option.
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The graph shows the relationship between the distance in miles and the time in minutes Scott ran during a race .
Answers
Answer:
Step-by-step explanation:
Scott ran 1 mile in 8 minutes
The slope of the line is 8 which means Scott ran 8 miles in 1-minute option (B) is correct.
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If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.
It is given that:
The graph of a line is shown in the picture
As we can see in the graph the line passes from the origin.
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Thus, the slope of the line is 8 which means Scott ran 8 miles in 1-minute option (B) is correct.
The complete question is mentioned in the picture.
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What is the 100th term in the sequence 11, 22, 33, 44, 55 …?
Answers
Its an arithmetic sequence. Here is the formula and process.
An= A1 + (n-1) d
An = the nth term in the sequence
A1 = 1st term
n = number of terms
d = common difference
An = A1 + (n-1) d
= 11 + (100-1) 11
= 11 + (99) 11
= 11 + 1,089
= 1,100
sana makatulongg study hard
The required 100th term in the sequence is given as 1100.
What is the 100th term in the sequences 11, 22, 33, 44, 55 … is to determine.
Arithmetic progression is the series of numbers that have common differences between preceding values.
Given sequence
11, 22, 33, 44, 55 …
Where first term ( a ) = 11 ;
common difference ( d ) = 11 - 22
= 11
The formula for the nth term of an Arithmetic progression is
a(nth) = a + ( n - 1 ) d
a (100th) = 11 + (100 - 1 )11
a (100th) = 11+ (99)11
= 11 + 1089
= 1100
The 100th term in an arithmetic sequence is 1100.
Thus, The required 100th term in the sequence is given as 1100.
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