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tino4ka555 [31]
3 years ago
8

Please help

Mathematics
1 answer:
Musya8 [376]3 years ago
5 0
(4,1) is the answer and the solution is in the image below

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If hexagon JGHFKI Is congruent to hexagon SRTQUV, which pair of angles must be congruent?
pav-90 [236]

Angles J and S or I and V would be congruent because they are corresponding.

3 0
4 years ago
Read 2 more answers
PLEASE HELP! <br> m∥n m∠1 = 50°, m∠2 = 48°, and line s bisects ∠ABC What is m∠3 ?
elena-14-01-66 [18.8K]

Answer:

m∠3 = 49°

Step-by-step explanation:

From the picture attached,

lines m and n are the parallel lines and line t is a transversal line intersecting these parallel lines at E and B respectively.

Therefore, ∠DEF ≅ ∠ABC [Exterior alternate angles]

m∠1 + m∠2 = m∠4 + m∠5

m∠4 = m∠5 [line s bisects ∠ABC]

50° + 48° = m∠4 + m∠4

98° = 2m∠4

m∠4 = 49°

Since, ∠4 ≅ ∠3 [Vertically opposite angles]

m∠3 = 49°

4 0
3 years ago
Help. Was not here today.
Elza [17]
If 205 is a posible answer I think thats's it
Because it's just like multiplucation 41x1=41
                                                        42x2=82.......etc.
                                                       42x5=205
6 0
3 years ago
The graphed line shown below is y=-3x+6
Bezzdna [24]
<h2>Answer:</h2>

y=-3\left(x+6\right)

<h2>Step-by-step explanation:</h2>

From the given graph, we can know two points:

P_{1}(x_{1},y_{1})=(2,0) \ and \ P_{2}(x_{2},y_{2})=(1,3)

Using the slope intercept form of the equation of a line, we get:

y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1}) \\ \\ y-0=\frac{3-0}{1-2}(x-2) \\ \\ y=-3(x-2) \\ \\ y=-3x+6

Where the slope is m=-3 and the y-intercept is b=6

From mathematics, we know that a system of two linear equations has no solution if and only if the two equations have the same slope, but not the same y-intercept. So, from the options, the second equation is:

y=-3(x+6)

And the slope intercept form of this equation is:

y=-3x-18

So the slope is m=-3 and the y-intercept is b=-18, the systems formed by these two equations has no solution.

6 0
4 years ago
Read 2 more answers
I need help I’ve been having trouble with this chapter for about a week
Harman [31]

Given:

3x^2+20x+33

Find-:

Factorization of the equation.

Sol:

A simple method of factorization is to multiply in first and last order then break it down into parts to make the middle number then.

\begin{gathered} =3\times33 \\ =99 \end{gathered}

The factor of 99 is:

So take factor :

\begin{gathered} 11\text{ and \lparen3}\times3) \\  \\ 11\text{ and  9} \end{gathered}

Factorization of the equation is:

\begin{gathered} =3x^2+20x+33 \\  \\ =3x^2+11x+9x+33 \end{gathered}

5 0
1 year ago
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