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Vika [28.1K]
3 years ago
15

What a addition problem that has a solution of -5

Mathematics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

-4+-1

Step-by-step explanation:

because a negative plus a negative is a negative.

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beks73 [17]

Answer:

A,C,D

Step-by-step explanation:

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2 years ago
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What is the difference between these two expressions?
mariarad [96]

Answer:

The answer is:

Option 4; (14+ 9) 4 is twice as much as(14 + 9) 2

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3 years ago
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9. A cup of tea costs x pence. Coffee costs 20 pence more. Hermes buys 2 cups of tea and one cup of coffee. In total Hermes pays
qaws [65]

Answer:

  • £1.20

Step-by-step explanation:

<u>Given:</u>

  • Tea = x
  • Coffee = x + 0.2

<u>Comparing the total cost, work out the value of x:</u>

  • 2x + x + 0.2 = 3.80
  • 3x = 3.60
  • x = 1.20
5 0
2 years ago
Points A, B, and C are collinear. Point B is between A and C. Solve for x if AC = 3x + 3, BC = 3, and AB = 2x + 2.
Ksenya-84 [330]

Answer:

x=2.

Step-by-step explanation:

It is given that points A, B, and C are collinear. Point B is between A and C.

Using segment addition property, we get

AC=AB+BC

It is given that AC = 3x + 3, BC = 3, and AB = 2x + 2.

3x+3=(2x+2)+3

3x+3=2x+5

Isolate variable terms.

3x-2x=5-3

x=2

Therefore, the value of x is 2.

5 0
3 years ago
Consider the differential equation x2y′′ − 9xy′ + 24y = 0; x4, x6, (0, [infinity]). Verify that the given functions form a funda
pantera1 [17]

Answer:

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

y= c_1x^4+c_2x^6

Step-by-step explanation:

Given equation is

x^2y'' - 9xy+24y=0

This Euler Cauchy type differential equation.

So, we can let

y=x^m

Differentiate with respect to x

y'= mx^{m-1}

Again differentiate with respect to x

y''= m(m-1)x^{m-2}

Putting the value of y, y' and y'' in the differential equation

x^2m(m-1) x^{m-2} - 9 x m x^{m-1}+24x^m=0

\Rightarrow m(m-1)x^m-9mx^m+24x^m=0

\Rightarrow m^2-m-9m+24=0

⇒m²-10m +24=0

⇒m²-6m -4m+24=0

⇒m(m-6)-4(m-6)=0

⇒(m-6)(m-4)=0

⇒m = 6,4

Therefore the auxiliary equation has two distinct and unequal root.

The general solution of this equation is

y_1(x)=x^4

and

y_2(x)=x^6

First we compute the Wronskian

W(x)= \left|\begin{array}{cc}y_1(x)&y_2(x)\\y'_1(x)&y'_2(x)\end{array}\right|

         = \left|\begin{array}{cc}x^4&x^6\\4x^3&6x^5\end{array}\right|

         =x⁴×6x⁵- x⁶×4x³    

        =6x⁹-4x⁹

        =2x⁹

       ≠0

The functions satisfy the differential equation and linearly independent since W(x)≠0

Therefore the general solution is

y= c_1x^4+c_2x^6

5 0
3 years ago
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