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Norma-Jean [14]
3 years ago
15

Help please asapppppppppppppppppppp

Mathematics
2 answers:
vovangra [49]3 years ago
7 0
The answer is 1.9! Explanation: 20.3-18.4=1.9
mylen [45]3 years ago
6 0
Yes that is correct
You might be interested in
The first line in a system of linear equations has a slope of 3 and passes through the point (-1 , -8). The second line passes t
gregori [183]
The equation of the first line can be written in point-slope form as
.. y = 3(x +1) -8
or
.. 3x -y = 5

The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6

A graph shows the solution to this system is (x, y) = (2, 1).

_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2

7 0
4 years ago
Can someone help me with these math problems please? I need it done by tonight if someone can please help me by then...I'll make
kotegsom [21]

Answer:

For question 3, you would just add 2 to the x values and subtract 2 from the y values, so it would be:

J' (-2, 5)

K' (2, 6)

L' (1, 2)

M' (-3, 1)

For question 4 you would subtract 7 from the x values and 6 from the y values, and that would be:

W' (-6, 1)

X' (-1, -1)

Y' (-3, -6)

Z' (-8, -4)

For question 9 you would end up with:

X' (6, -5)

Y' (7, 1)

Z' (4, 0)

For question 10 you would end up with:

Q' (-1, 2)

R' (1, 7)

S' (-2, 6)

T' (-4, 1)

For question 11 you would end up with:

L' (4, 1)

M' (8, 5)

N' (6, 7)

P' (2, 3)

For question 12 you would end up with:

G' (6, -7)

H' (6, -4)

I' (1, -7)

Hope this is what you were looking for!

 

Step-by-step explanation:


4 0
3 years ago
In triangle JKL, m of angle J = 90, m of angle K = 30, and m of angle L = 60. Which of the following statements about triangle J
Lynna [10]

Answer:

option C. KL=2*JL

option D. JK=\frac{\sqrt{3}}{2}*KL

option F. JK=\sqrt{3}*JL

Step-by-step explanation:

step 1

we know that

In the right triangle JKL

sin(30\°)=\frac{JL}{KL}

sin(30\°)=\frac{1}{2}

so

\frac{1}{2}=\frac{JL}{KL}

KL=2*JL

step 2

cos(30\°)=\frac{JK}{KL}

cos(30\°)=\frac{\sqrt{3}}{2}

so

\frac{\sqrt{3}}{2}=\frac{JK}{KL}

JK=\frac{\sqrt{3}}{2}*KL

step 3

tan(30\°)=\frac{JL}{JK}

tan(30\°)=\frac{1}{\sqrt{3}}

\frac{1}{\sqrt{3}}=\frac{JL}{JK}

JK=\sqrt{3}*JL

7 0
4 years ago
Please solve needed as soon as possible thank you I really appreciate it
e-lub [12.9K]

Answer:

143

Step-by-step explanation:

Substitute n=3 into the problem

6(3^{3})-2(3^{2}) +3-4

Then figure out squared and power by 3

6(27)-2(9)+3-4

Next multiply

162-18+3-4

Finally, add and subtract

144+3=147

147-4=143

8 0
3 years ago
Jack's mother gave him 50 chocolates to give to his friends at his birthday party. He gave 3 chocolates to each of his friends a
kakasveta [241]

Answer:

50/3=16

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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