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3 years ago

How to do the problem I don't get it

Mathematics

1 answer:

Ad libitum [116K]3 years ago

5 0

The rate of change=250-125/2-1
=125
the initial value is 700

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Prove algebraically that (m + 2)2 – m 2 – 12 is always a multiple of 4

Julli [10]

Answer:

$(m + 2)^{2} - m^{2} - 12 = 4(m - 2)$

Step-by-step explanation:

Step 1:

Write the expression

$(m+2)^{2} - m^{2} - 12$

Step 2: Expand $(m + 2)^{2}$

$(m+2)^{2} - m^{2} - 12\\(m+2)(m+2) - m^{2} - 12\\m^{2} + 2m + 2m + 4 - m^{2} - 12$

Step 3: Collect similar terms

$m^{2} - m^{2} + 4m + 4 - 12\\4m - 8$

Step 4: Factor 4 out of the expression to prove that the expression is a multiple of 4.

$Therefore\\4m - 8 = 4(m - 2)\\Hence,\\(m+2)^{2} - m^{2} - 12 is a multiply of 4 because the expression is equal to 4(m-2)$

3 0

2 years ago

Consider the two triangles.

GuDViN [60]

Answer:

(A)Show that the ratios StartFraction U V Over X Y EndFraction , StartFraction W U Over Z X EndFraction , and StartFraction W V Over Z Y EndFraction are equivalent.

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

Step-by-step explanation:

In Triangles WUV and XZY:

$\angle VUW$ and \angle YXZ$ are congruent. \\\angle U W V$ and \angle X Z Y$ are congruent.\\ \angle U V W$ and \angle Z Y X$ are congruent.$

Therefore:

$\triangle UWV \cong \triangle XZY$

To show that the triangles are similar by the SSS similarity theorem, we have:

$\dfrac{UW}{XZ}=\dfrac{WV}{ZY}=\dfrac{UV}{XY}$

As a check:

$\dfrac{UW}{XZ}=\dfrac{40}{32}=1.25\\\\\dfrac{WV}{ZY}=\dfrac{60}{48}=1.25\\\\\dfrac{UV}{XY}=\dfrac{50}{40}=1.25$

The correct option is A.

4 0

3 years ago

Hey! I would really love it if someone could help me

Umnica [9.8K]

Alrigth and B....

I think??????????? If I get it wrong im sorry ;-;

4 0

2 years ago

Fabio and Carlos play on a basketball team together. In the last game, Fabio had 7 points less than 2 times as many points as Ca

Ymorist [56]

Let 'c' represent the number of points Carlos scored. 2c - 7 = 31 c = 19 points

7 0

3 years ago

Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.

liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

$f(x) = \frac{x^2+4x+3}{x^2-x-2}$

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial $x^2-x-2$

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

$(x-2) (x + 1)$

We do the same with the numerator

$x^2+4x+3$

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

$(x+3)(x + 1)$

Therefore

$f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}$

Note that $\frac{(x+1)}{(x+1)}=1$ only if $x \neq -1$

So since $x = -1$ is not included in the domain the function has a discontinuity in $x = -1$

3 0

3 years ago