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CaHeK987 [17]
3 years ago
8

Solve log base 2 of one over thirty two.

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
8 0
It’s in this picture:)

lisov135 [29]3 years ago
4 0

Answer:

Step-by-step explanation:

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somebody help pls Complete the table by filling in the multiplicative inverse of the numbers below 2,1/5,-4,​
Yakvenalex [24]

Answer:

Suppose we have a random number A.

The multiplicative inverse of A is a number X such that:

A*X = 1

When we work with real numbers, X = 1/A

Then:

A*(1/A) = A/A = 1

This means that (1/A) is the multiplicative inverse of A.

Where we need to have A ≠ 0, because we can not divide by 0.

Now we want to find the multiplicative inverse of the numbers:

2: Here the inverse is (1/2) = 0.5

1/5:  Here the inverse is  (1/(1/5)) = (5/1) = 5

-4: Herre the inverse is  (1/(-4)) = -(1/4) = -0.25

7 0
3 years ago
Find the solution to the initial value problem
Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0
3 years ago
Simplify f(x)=(x+1)(x-3) to standard form
Ivahew [28]

Answer:

x^2-2x-3

Explanation: Use FOIL

4 0
4 years ago
Two numbers are 10 units away in different directions from their midpoint, m, on a number line. The product of the numbers is –9
irinina [24]
In my opinion the answer is <span>m2 – 100 = –99
proof
</span>m2 – 100 =<span>(m – 10)(m + 10)
for eg. m=1, </span>(-9)(11)= <span>–99</span>
4 0
3 years ago
Read 2 more answers
WILL MARK YOU BRAINLIEST !!!!
liq [111]

Answer:

\overline {PM} \cong \overline {ON} because they are opposite sides of a parallelogram.

Step-by-step explanation:

First of all, let us have a look at the definition of a parallelogram.

<em>A parallelogram is a closed 4 sided figure (i.e. a quadrilateral) made up with two pairs of straight lines such that the two pairs are parallel and equal to each other.</em>

<em></em>

For example, let us consider a quadrilateral ABCD as attached in the diagram in answer area.

The two pair of lines are:

AB,  DC and BC, AD

For ABCD to be a parallelogram, the lines

AB and DC must be parallel to each other and AB = DC

AND

BC and AD must be parallel to each other and BC = DA

------------------------------

In the given question, we are given that the quadrilateral MNOP is a parallelogram.

The two pairs opposite lines are OP, NM and PM, ON.

As per the definition of a parallelogram,

OP and NM must be <em>parallel and equal </em>to each other.

AND

PM and ON must be <em>parallel and equal </em>to each other.

\therefore \overline {PM} \cong \overline {ON}.

Hence proved.

6 0
3 years ago
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