solve for p if 2|p|=4

Answer:

false 9.13 is greater

Step-by-step explanation:

103 thousanths are smaller then 13 hundreths

with decimals the farther you go the lower you get bc its half a number

1/2 to 1/4 to  1/8 it gets smaller

Answer:

42 125 becuase i got um right bozzzzzzzz

Step-by-step explanation:

Answer:

introducing customer retention strategies

Step-by-step explanation:

-8,-6,-4

-10,-6,0

these are the possible test points for your three intervals

The given inequality is, x² + 12 x+35≥0

→ x²+7 x +5 x+5×7≥0

→ x (x+7) +5(x+7)≥0

→ (x+5)(x+7)≥0

→ x+5=0  ∧ x+7=0 gives x= -7 ∧ x= -5

Now drawing the number line and marking point ,-5 and -7 on it.

Now the three intervals are (-∞ , -7], [-7,-5] and [-5,∞).

The solution set of inequality (x+5)(x+7)≥0 is  (-∞ , -7] and [-5,∞).

The set of possible test points for

⇒ (-∞ , -7] → -8,  -10

⇒ [-7,-5] → -6

⇒ [-5,∞) → -4, 0

Option (1) -8,-6,-4 and Option (2) -10,-6,0 satisfies the given condition.

Final answer:

To find two consecutive even integers, we can set up and solve an equation. The equation in this case is an identity, meaning it is true for all values of x. Therefore, there are infinitely many solutions.

Explanation:

To find two consecutive even integers, let's represent the first integer as x and the second integer as x + 2. Now, according to the given information, twice the lesser of the two integers is 4 less than two times the greater integer. We can write this as:

2x = 2(x + 2) - 4

Simplifying the equation:

2x = 2x + 4 - 4

2x = 2x

This equation is an identity, meaning it is true for all values of x. Therefore, there are infinitely many solutions. In this case, there are infinitely many pairs of consecutive even integers that satisfy the given conditions.

The domain of the composite function (cd)(x), where c(x) = 5/(x-2) and d(x) = x + 3, includes all real values except x = 2 due to a division by zero condition in the function c(x).

In this problem, we need to figure out the domain of the combined function (cd)(x). The domain constitutes the set of all real values of x that make the function defined. The first function, c(x) =  rac{5}{x-2}, isn't defined for x = 2 because it would lead to a division by zero. The second function, d(x) = x + 3, is defined for all real values since it's a linear function.

When these functions are multiplied to form (cd)(x), the domain will be the intersection of the individual function domains. This means, with c(x) having a restriction with x != 2, the composite function (cd)(x) will maintain this restriction despite d(x) being defined for all real values. So the domain for (cd)(x) is all real values except for x = 2.

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I KNOW ITS NOT A. But i think it is B. HOPE THIS HELPS!

Answer:

option B  

Step-by-step explanation:

Given in the question are ∠x ≅ ∠y and ∠zwx ≅ ∠ ywz and wz is common in both.

These angles are adjacent angles of triangles.

We know theorem for congruence is Angle - Angle - Side

Therefore, for the congruence of these two triangles option B, AAS congruence theorem is the answer.

Answer:

Option A, C, D, are the correct options.

Step-by-step explanation:

To check the exponential functions which have been simplified correctly we will take option one by one.

A). [tex]f(x)=5(\sqrt[3]{16})^{x}=5(\sqrt[3]{2^{4}})^{x}=5(2\sqrt[3]{2})^{x}[/tex]

B). [tex]f(x)=2.3(8^{\frac{1}{2}})^{x}=2.3(2.\sqrt{2})^{x}\neq 2.3(4^{x})[/tex]

C). [tex]f(x)=81^{\frac{x}{4}}=(3^{4})^{\frac{x}{4}}=(3^{\frac{4}{4}})^{x}=3^{x}[/tex]

D). [tex]f(x)=\frac{3}{4}(\sqrt{27})^{x}=\frac{3}{4}(\sqrt{3^{3}})^{x}=\frac{3}{4}.(3\sqrt{3})^{x}[/tex]

[tex]f(x)=(24)^{\frac{1}{3}x}=(2^{3}.3)^{\frac{1}{3}x}=(2.3^{\frac{1}{3}})^{x}\neq 2(\sqrt[3]{3})^{x}[/tex]

Exponential functions and their inverses like the natural logarithm are inverse functions, meaning they can undo each other. Simplifications like ln(e^x) result in x, and e^(ln x) simplifies to x. This property is integral to correctly simplifying exponential functions.

The exponential functions and their inverses, like the natural logarithm (ln), have specific properties where they can "undo" each other. This relationship allows us to simplify expressions involving exponentials and logarithms. For example, when you have ln(ex), this simplifies to x because the natural logarithm and the exponential function are inverse functions.

Similarly, when applying the exponential function to a natural logarithm, like eln(x), the result is just x. Furthermore, when taking square roots of exponentials, you can adjust the exponent to be evenly divisible by 2 and then take the square root of the base number, followed by dividing the exponent by 2. Also, while dividing exponentials, you subtract the exponents when the bases are the same.

Here is an example of this simplification: to compute the square root of e4, we can write it as e4/2, which simplifies to e2.

Answer:

Another word for mean is the "average", your goal is to find the average of the following data set, it's pretty easy, you just have to add up all the numbers in the data set and divide it with the total numbers in the data set.

For example:

Find the mean of the following data set:

4, 1, 8, 5, 7, 3

Mean: 4 + 1 + 8 + 5 + 7 + 3 = 28

then we divide 28 with 6 since there are 6 numbers in the following data set.

Mean: 4 + 1 + 8 + 5 + 7 3 = 28/6 = 4.6

We get the final answer of 4.6!

The postulate that proves that △JHG and △EFG are congruent is;

Option B; ​ ASA Congruence Postulate ​

From the given image, we can see two triangles namely;

△JHG and △EFG

Now, we can see that;

FG = GH

That is given because we see that G has divided FH into 2 equal parts.

Also, we see the right angle sign at ∠F and ∠H of both triangles.

Thus, since they are both right angles, then we can say that;

∠EFG = ∠JHG

Also, at point G, we have two vertically opposite angles which are;

∠EGF and ∠JGH

Now, from vertically opposite angle theorem, vertically opposite angles are always equal;

Thus; ∠EGF = ∠JGH

So far, we have;

FG = GH

∠EFG = ∠JHG

∠EGF = ∠JGH

This is 2 corresponding equal angles and 1 corresponding equal side. The side is included between the two angles and as such the proof that  △JHG and △EFG are congruent is the ASA Congruence Postulate.

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Answer:

43.755 what the other person said but I know that is the answer

Step-by-step explanation: