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Evgesh-ka [11]
2 years ago
12

Use substitution to solve the following system of equations. What is the value of y?

Mathematics
1 answer:
koban [17]2 years ago
5 0
The answer is C. I solved for the first variable in one of the equations then substitute the result into the other equation.
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(3x4)^4 simplify using power rule
hichkok12 [17]

Answer:

12 to the fourth power

4 0
3 years ago
Solve for x by first setting up two equations. Then find the PRODUCT of your solutions |2x+5|=7
Kobotan [32]

Answer: see below

Step-by-step explanation:

|2x+5|=7

2x+5=±7 ⇔ since the absolutely value of any number is all positive, so the value in side can be either positive or negative

2x+5=7  or  2x+5=-7

    2x=2            2x=-12

      x=1      or       x=-6

----------------------------------------

The product of the two solutions

1 × -6=-6

Hope this helps!! :)

Please let me know if you have any question

5 0
3 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
3 years ago
Read 2 more answers
You are remodeling your kitchen. You've contacted two tiling companies who gladly told you how long it took their workers to til
quester [9]
Ok so pete lays 10 more than jim per hour

pete and jim laid same amount of tiles

j=jim's rate
p=pete's rate

6j=5p

but pete lays 10 tile per hour faster so
p=10+j

sub
6j=5(10+j)
6j=50+5j
minus 5j both sides
j=50


jim can lay at 50 tiles per hour
7 0
3 years ago
Read 2 more answers
find the area of a circle that has a diameter of 11inches approximate as 13.4 round your answer to the hundreth
Vinil7 [7]

To find the area of a circle you do pi (3.14) multiplied by the radius squared.

The radius is half of the diameter: 11 / 2 = 5.5in

The radius squared is 5.5^2 = 30.25

Then take the squared radius and multiply that by pi 30.25 * 3.14 = 94.985

So the area of the circle is about 94. 99 in.

Hope this helped! Mark as Brainliest please! :)))

3 0
3 years ago
Read 2 more answers
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