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sergij07 [2.7K]
3 years ago
8

Which statement can john use this figure to prove

Mathematics
1 answer:
Brums [2.3K]3 years ago
3 0

Answer:

The correct option is;

The points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects

Step-by-step explanation:

With the assumption that that the question meant that ΔACD and ΔBCD are congruent, we have;

Segment \overline {AC} is congruent to segment \overline {BC} by Congruent Parts of Congruent Triangles are Congruent, CPCTC

Similarly, segment \overline {DA} is congruent to segment \overline {DB} by Congruent Parts of Congruent Triangles are Congruent, CPCTC

Therefore, the points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects

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If y varies directly as x and y = 4 and x = 10, what does x equal when y = 7?
soldi70 [24.7K]

Answer: x = 17 \frac{1}{2}

Step-by-step explanation:

From the question: y α x

⇒ y = kx

y=4 and x = 10 implies

4 = 10k

k = 4/10

k = 2/5

substitute k = 2/5 into the equation

y = 2/5 x

When y = 7 , the equation becomes

7 = \frac{2x}{5}

35 = 2x

35/2 =x

Therefore : 17 \frac{1}{2}

x =

8 0
4 years ago
Write the equation of the line with a slope of 2 and passes throug the point (4, 3).
Igoryamba

Answer:

y=2x-5

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Identify the value of x and the length of each secant segment. HELP PLS options: x = 15; PR = 12; PT = 19 x = 15; PR = 19; PT =
Yakvenalex [24]

Answer:

Option C.

Step-by-step explanation:

From the given figure it is clear that

PQ=5,QR=7,PS=4,ST=x

So,

PR=PQ+QP=5+7=12

PT=PS+ST=4+x

Using Intersecting Secants Theorem, we get

PQ\times PR=PS\times PT

5\times 12=4\times (4+x)

60=16+4x

60-16+4x

44=4x

Divide both sides by 4.

11=x

PT=4+x=4+11=15

Since, x = 11; PR = 12; PT = 15, therefore the correct option is C.

8 0
3 years ago
A skier has decided that on each trip down a slope, she will do 2 more jumps than before. On her first trip she did 6 jumps. Der
elixir [45]

The <em>correct answers</em> are:

\Sigma_{n=4}^{12} 6+2(n-1) \\ \\=180

Explanation:

Since she is adding two more jumps every time she goes down hill, this is an arithmetic sequence. The general form of an arithmetic sequence is

a_n=a_1+d(n-1), \\ \text{where } a_n \text{represents the nth term, } a_1 \text{represents the first term, and n represents the term number}

Since we want the number of jumps on her 4th through 12th trips, we will set n in the summation from 4 to 12. n=4 goes at the bottom of Σ and 12 goes at the top, to represent the values we are interested in.

Beside this, we write our general form. The first term is 6 and d, the common difference, is 2. This gives us 6+2(n-1) beside the summation:

\Sigma_{n=4}^{12} 6+2(n-1)

To evaluate this, we substitute the values 4, 5, 6, 7, 8, 9, 10, 11 and 12 in for n, adding all of the values together:

6+2(4-1)+6+2(5-1)+6+2(6-1)+6+2(7-1)+6+2(8-1)+6+2(9-1)+6+2(10-1)+6+2(11-1)+6+2(12-1)

=6+6+6+8+6+10+6+12+6+14+6+16+6+18+6+20+6+22

=180

8 0
3 years ago
Read 2 more answers
Prove diagonals of a rhombus are perpendicular
Lisa [10]

Answer:

A rhombus is a parallelogram with four congruent sides.

So, all sides of rhombus ABCD are congruent.

i.e, \overline{JM} \cong \overline{JK} \cong \overline{KL}\cong \overline{LM}

Also, we know that the diagonals of a parallelogram bisect each other.  

Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

\overline{MN} \cong \overline{NK}         .....[1]

\overline{JN} \cong \overline{NL}

In ΔJNM and ΔJNK

\overline{MN} \cong \overline{NK}     [side]    [by (1)]

\overline{JM} \cong \overline{JK}     [side]          [Given]

By reflexive property states that a segment is congruent to itself:

\overline{JN} \cong \overline{JN}   [Side]                 [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

\triangle JNM \cong \triangle JNK

By CPCT [Corresponding Part of congruent triangles are congruent]

Since, JNM and JNK are corresponding angles therefore,

\angle JNM \cong JNK

Linear pair theorem states that two angles that form a linear pair are supplementary.

By linear pair theorem, JNM and JNK are supplementary

this mean:

m\angle JNM +m\angle JNK =180^{\circ}

Since, the angles are congruent i.e, \angle JNM \cong JNK

so;

m\angle JNK +m\angle JNK =180^{\circ}

or

m\angle JNK +m\angle JNK =180^{\circ}

2m\angle JNK=180^{\circ}

Simplify:

m\angle JNK =90^{\circ}

also; m\angle JNM =90^{\circ}

therefore, the diagonals of JKLM are perpendicular to each other i,e \overline{JL} \perp \overline{MK}


4 0
3 years ago
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