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Jobisdone [24]
2 years ago
8

Please help with the question attached, will mark brainliest!!!!!!!!!!!

Mathematics
1 answer:
UkoKoshka [18]2 years ago
7 0

12; 4

66; 33.4

32.6

respectively in the blanks

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What's the pattern for 2,4,7,9
Semenov [28]

Answer:

The pattern is this: I create a function p(x) such that

p(1)=1

p(2)=1

p(3)=3

p(4)=4

p(5)=6

p(6)=7

p(7)=9

Therefore, trivially evaluating at x=8 gives:

p(8)= 420+(cos(15))^3 -(arccsc(0.304))^(e^56) + zeta(2)

Ok, I know this isn’t what you were looking for. Be careful, you must specify what type of pattern is needed, because the above satisfies the given constraints.

Step-by-step explanation:

4 0
3 years ago
What is 32.330 + 23.559
Molodets [167]
I hope this helps you

5 0
3 years ago
A student provided the following solution for an equation. What mistake did the student make?
kvv77 [185]

Answer:

1st choice - 1st line- Distributive property applied incorrectly

Step-by-step explanation:

3(a-2) should be 3a-6

3 0
2 years ago
HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME
Kryger [21]

Answer:

1)

Given the triangle RST with Coordinates  R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.  

Since, the scale factor \frac{5}{3} is greater than 1, the image is enlargement or a stretch.  

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor \frac{5}{3}

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image S{}' are same.

Now, the distances from the center of the dilation at point S to the other points R and T.  

The dilation image will be\frac{5}{3} of each of these distances,

SR=3, so S{}'R{}'=5 ;


ST=3, so S{}'T{}'=5  

Now, draw the image of RST i.e R'S'T'

Since, RT=3\sqrt{2} [By using hypotenuse of right angle triangle] and R{}'T{}'=5\sqrt{2}.


2)

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of \overline{AC} i.e \overline{AB}\cong \overline{BC}

In the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

Since, there is no included angle in these triangles.

∴ \Delta ABD is not congruent to \Delta CBD .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

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

Since it is also given that  \overline{AD}\cong \overline{CD}.

therefore, in the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{AD}\cong \overline{CD}   (SIDE)           [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

therefore by, SSS postulates \Delta ABD\cong \Delta CBD.

3)

Given that:  \angle1=\angle 3 are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

\angle 1 and \angle 2 and \angle 3  and \angle 2 are linear pair.

By linear pair theorem, \angle 1 and \angle 2   are supplementary, \angle 2 and \angle 3  are supplementary.

m\angle1+m\angle 2=180^{\circ}

m\angle2+m\angle 3=180^{\circ}

Equate the above expressions:

m\angle 1+m\angle 2=m\angle 2+m\angle 3

Subtract the angle 2 from both sides in the above expressions

∴m\angle 1=m\angle 3

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.


therefore, \angle 1\cong \angle 3.















3 0
2 years ago
Read 2 more answers
a computer and printer a totoal cost $1132 the cost of the computer is three times the cost of the ptinter
zavuch27 [327]
Use a system of equations

C+P=1132
3P=C

Substitute C in first equation as
3P+P=1132
Simplify
4P=1132
Solve
P=1132/4
P=283

NOW SOLVE FOR C SUBSTITUTING P VALUE IN FIRST EQUATION

C+283=1132
C=1132-283
C=849


Printer = 283$
Computer = 849$
3 0
2 years ago
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