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Tasya [4]
3 years ago
7

Please help SUMMER SCHOOL

Mathematics
1 answer:
mote1985 [20]3 years ago
7 0
I am not sure what you mean by slope but the y intercept is (0,2) and the gradient is 3 making the equation y=3x+2
Explanation:
The y-intercept is where the line crosses the y axis and I have put the co-ordinate where it does that.
The gradient is 3 as to find the gradient you do rise/run so we do 3/1 which equals 3
To form an equation we do y=Mx+c. The c is the y-intercept and the gradient is m. Therefore the equation of this line is y= 3x+2.
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please solve the question and show your work find the missing number in 34x =0 show how u got your answer and write full explana
Anastaziya [24]
34x=0
So we need to isolate x. Since 34 is multiplied by x, we need to isolate x by moving 34 over the equal sign. Whenever you move something over the equal sign, the opposite is done. For 34, instead of being multiplied, it will be divided.
34x=0
X=0/34
Then you solve for x
X=0
Hope this helps!
8 0
3 years ago
A teacher is four times as old as a student. In 28 years, the student's age will be half of the teacher's age. How old are they
swat32
Here, let the age of student be x and teacher be 4x after 28 years, the teacher is double the age of student. So the age of teacher will be 4x+28 and student will be 2(x+28)
So,
4x+28=2(x+28)
4x+28=2x+56
4x-2x=56-28
2x=28
x=14
Therefore the age of the teacher in present will be
=4x+28
=4*14+28
=84
And age of student will be:
=x+28
=14+28
=42
Hope this helps


7 0
3 years ago
Find the jacobian of the transformation. x = u2 + uv, y = uv2
atroni [7]
\mathbf J=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{bmatrix}=\begin{bmatrix}2u+v&u\\v^2&2uv\end{bmatrix}

The Jacobian has determinant

\det\mathbf J=\begin{vmatrix}2u+v&u\\v^2&2uv\end{vmatrix}=(2u+v)2uv-uv^2=4u^2v+uv^2=uv(4u+v)
5 0
4 years ago
Solve for m.
elixir [45]
M-7/8=5
m=5+7/8
m=47/8

Fraction: 47/8
Mixed Fraction: 5 7/8
4 0
3 years ago
Read 2 more answers
Corbin made a scale model of the San Jacinto Monument. The monument has an actual height of 604 feet. Corbins model used a scale
sergejj [24]

Answer: 6.04 inches


Step-by-step explanation:

Given: The actual height of the monument = 604 feet

The scale used by Corbins  is 1 inch = 100 feet

It implies in Corbins model 1 feet =\frac{1}{100}\ inches

Therefore, The height of the Corbins model in inches= \frac{604}{100}=6.04\ inches

Hence, the height of the Corbins model in inches= 6.04 inches

8 0
3 years ago
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