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arsen [322]
3 years ago
10

Can you help me on these math problems?

Mathematics
1 answer:
densk [106]3 years ago
8 0
I cannot help you unless I know what ones you need help on
You might be interested in
For which values of P and Q does the following equation have infinitely many solutions? Px-37=Qx-37
Lyrx [107]

Answer:

The equation has infinitely many solutions for any value of P and Q such that P=Q.

Step-by-step explanation:

Px - 37 = Qx - 37

Px - Qx - 37 = -37

x (P-Q) = 0

==> P-Q=0 ==> P=Q


4 0
3 years ago
Complete the statement, f( ) = -5.<br><br> Blank 1: <br> Plz answer asap!
tankabanditka [31]
It’s 12 that’s the simple answer
4 0
3 years ago
Tell whether you can make a unique triangle with sides measuring 5cm, 4cm, and 20cm.​
Marina CMI [18]

Answer:

The answer your looking for is no, because in any certain type of triangle. Each triangle is 180 degrees like trying to find the missing angle of a triangle.

​

Step-by-step explanation:

6 0
3 years ago
Write an equation for a line that is parallel to 2x+5y=15 and passes through the point (-10, 3)​
Lyrx [107]

Answer:

The equation of the line is,

y =  -  \frac{2}{5} x - 1

Step-by-step explanation:

First, you have to write it in a form of y = mx + b :

2x + 5y = 15

5y = 15 - 2x

y = 3 -  \frac{2}{5} x

y =   - \frac{2}{3} x + 5

When both lines are parallel to each other, they will have to same gradient value. So the equation of the line is y = (-2/5)x + b. Next, you have to find the value of b by substutituting (-10,3) into the equation :

y =   - \frac{ 2}{5}x + b

let \: x =  - 10,y = 3

3 =  -  \frac{2}{5} ( - 10) + b

3 = 4 + b

3 - 4 = b

b =  - 1

3 0
3 years ago
Read 2 more answers
Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 2x − 4x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
bekas [8.4K]

9514 1404 393

Answer:

  • (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
  • NOT linearly independent

Step-by-step explanation:

We want ...

  c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0

Substituting for the fn function values, we have ...

  c1·x +c2·x² +c3·(2x -4x²) ≡ 0

This resolves to two equations:

  x(c1 +2c3) = 0

  x²(c2 -4c3) = 0

These have an infinite set of solutions:

  c1 = -2c3

  c2 = 4c3

Then for any parameter t, including the "trivial" t=0, ...

  (c1, c2, c3) = (-2t, 4t, t)

__

f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)

7 0
2 years ago
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