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VikaD [51]
3 years ago
14

Fill in the missing value to make the equations true.

Mathematics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

a) 7/4 b) 5 c) 2

Step-by-step explanation:

Logrithmic Rule for a and b

Let a, M, N be positive real numbers.

a)  

logaM - logaN = loga(M/N)

log9(7) - log9(4) = log9 (7/4)

b)

logaM + logaN = logaMN

log2 (x) + log2(9) = log2(45)

x9=45

(x9)/9 = 45/9

x = 5

c)

Change of base formula.

logb​(x)=logd​(b)/logd​(x)​

x log6(5) = log6(25) divide each term by log6(5)

x log6(5) / log6(5) = log6(25) / log6(5)   Cancel common factor log6(5)

x = log6(25) / log6(5)

x = log6(5^2) / log6(5)

Expand log6(5^2) by moving 2 outside the logarithm.

x = 2log6(5) / log6(5) cancel the like term log6(5)

x = 2

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mote1985 [20]

Answer:

x=1

y=-4

Step-by-step explanation:

4x+y=0  ->    y= -4x

2(-4x)+x=-7

-8x+  x=-7

-7x=-7

x=-7/-7=1

y=-4 . 1 = -4

3 0
3 years ago
A- CE=CD <br> B- CE = CA <br> C- BF=DF <br> D- DF=EF<br><br> please help!!
Oksana_A [137]

The perpendicular bisector theorem gives the statements that ensures

that \overleftrightarrow{FG} and \overleftrightarrow{AB} are perpendicular.

The two statements if true that guarantee  \overleftrightarrow{FG} is perpendicular to line \overleftrightarrow{AB} are;

  • \overline{CE} = \overline{CD}
  • \overline{DF} = \overline{EF}

Reasons:

The given diagram is the construction of the line \mathbf{\overleftrightarrow{FG}} perpendicular to line \mathbf{\overleftrightarrow{AB}}.

Required:

The two statements that guarantee that  \overleftrightarrow{FG} is perpendicular to line \overleftrightarrow{AB}.

Solution:

From the point <em>C</em> arcs <em>E</em> and <em>D</em> are drawn to cross line \overleftrightarrow{AB}, therefore;

\overline{CE} = \mathbf{\overline{CD}} arcs drawn from the same radius.

\overleftrightarrow{FG} is perpendicular to line \overleftrightarrow{AB}, given.

Therefore;

\overline{DF} = \overline{EF}  by perpendicular bisector theorem.

Learn more about the perpendicular bisector theorem here:

brainly.com/question/11357763

7 0
2 years ago
Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k
Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

7 0
3 years ago
Simplify 3^2/3.3^1/5​
san4es73 [151]

ANSWER:

\frac{1}{5}  \\

EXPLANATION:

\frac{ \frac{{3}^{2}}{3 \times{3}^{1}}}{5}  =  \\

\frac{ \frac{9}{3 \times 3} }{5}  =  \\

\frac{ \frac{9}{9} }{5}  =  \\

\frac{1}{5}  \\

4 0
3 years ago
The graph of a function f(x) is shown below:<br><br><br><br> What is the domain of f(x)?
Pie

The domain is the scope of the x values. In this case (-2,-1) is a hole, so x>-2 is one end of the domain, while (3,3) is defined. So the domain using interval notation is (-2,3] which can also be expressed -2 < x ≤ 3, answer option 1

8 0
3 years ago
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