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ValentinkaMS [17]

2 years ago

10

Consider the following number line:

2 answers:

JulijaS [17]2 years ago

7 0

Answer:

The pink A goes to the Green C on the number line. the green C goes to the pink A on the number line. The Blue B goes to the Blue B in the number line.

Step-by-step explanation:

hope it helped ;)

kirza4 [7]2 years ago

7 0

The pink somthing what the persons above me said

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Pls help to get the answer

soldier1979 [14.2K]

When they say quotient, they want a fraction, so the answer for that would be
$\frac{6}{10}$
and the decimal form of that would be 0.6. You can get that from dividing 6 by 10 and solving with long division.

3 0

3 years ago

An amusement park charges $5 for admission and $.80 for each ride. You go to the park with $13. Which inequality represents this

RSB [31]

Answer:

Answer Placeholder (I'm going to solve it and answer when I'm done)

Step-by-step explanation:

7 0

3 years ago

Soccer balls cost $24 each at Sports Emporium. Coach Neville can spend at the most $120 on equipment for the soccer team. Let b

MrRissso [65]

Answer:

5

Step-by-step explanation:

24b≤120

3 0

2 years ago

A,B and C are the vertices of one triangle.

dsp73

Answer:

Step-by-step explanation:

Given: The triangle with coordinate A(4,6), B(2,-2) and C(-2,-4). D is the mid point of AB and E is the mid point of AC.

To prove: DE is parallel to BC.

Construction: Join DE.

Proof: If we prove the basic proportionality theorem that is $\frac{AD}{DB}=\frac{AE}{EC}$ , then it proves that DE is parallel to BC.

Now, Mid Point D has coordinates= $(\frac{4+2}{2},\frac{6-2}{2})=(3,2)$ and Mid Point E has coordinates= $(\frac{4-2}{2},\frac{6-4}{2})=(1,1)$

Now, AD= $\sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17}$

DB= $\sqrt{(3-2)^{2}+(2+2)^{2}}=\sqrt{17}$

AE= $\sqrt{(4-1)^{2}+(6-1)^{2}}=\sqrt{34}$

EC= $\sqrt{(1+2)^{2}+(1+4)^{2}}=\sqrt{34}$

Now, $\frac{AD}{DB}=\frac{AE}{EC}$

= $\frac{\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{34}}{\sqrt{34}}=\frac{1}{1}$

Hence, $\frac{AD}{DB}=\frac{AE}{EC}$

Thus, By basic proportionality theorem, DE is parallel to BC.

4 0

3 years ago

Read 2 more answers

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

Serga [27]

Answer:

Required solution is $y(x)=1+A\cos x+B\sin x$ where A and B are constants.

Step-by-step explanation:

Given nonhomogenous differential equation is,

$y''(x)+y'(x)=1\hfill (1)$ with $y_1(x)=1$

To find another solution, consider $m=\frac{\partial}{\partial x}$ be such that,

$m^2+1=0\implies m=\pm i$

Hence,

$C.F=A\cos x+B\sin x$ where A and B are constants.

Let $D=\frac{\partial}{\partial x}$

$P.I=\frac{1}{1+D^2}(1)$

$=(1+D^2)^{-1}(1)$

$=(1-D^2+.....)(1)=1$

Hence,

$y(x)=1+A\cos x+B\sin x$ where A and B are constants.

which is required solution.

6 0

3 years ago