Solve 3/10 n-3 1/5 = -1 1/5

Answers-
A.n =-12 2/3
B.n = 6 2/3
C.n = 8
D.n = -6/25

Answer:

  I agree with Sydneyjones :D i very much agree

1.) 32 start root 2 end root

2.) -5 start root 5 end root

3.) 3 start root 7 end root + 7 start root 6 end root

4.) start root 10 end root + start root 6 end root

The simplified form of the expression: [tex]\( -\frac{15}{2^{24} \times 5^{12} \times p^4 \times q^3} \)[/tex]

To simplify the expression [tex]\( \frac{15p^{-4}q^{-6}}{-20^{12}q^{-3}} \)[/tex], we can apply the rules of exponents and simplify each part of the expression separately.

First, let's rewrite [tex]\( -20^{12} \) as \( -(20^{12}) \)[/tex] to make it clear that the exponent applies to the entire term:

[tex]\[ \frac{15p^{-4}q^{-6}}{-(20^{12}q^{-3})} \][/tex]

Now, let's simplify each part:

1. \( p^{-4} \) can be rewritten as [tex]\( \frac{1}{p^4} \).[/tex]

2. \( q^{-6} \) can be rewritten as [tex]\( \frac{1}{q^6} \).[/tex]

3. [tex]\( 20^{12} \)[/tex] can be calculated.

[tex]\[ \frac{15}{-(20^{12})} \times \frac{1}{p^4} \times \frac{1}{q^6} \times \frac{1}{q^{-3}} \][/tex]

Now, let's calculate [tex]\( 20^{12} \):[/tex]

[tex]\[ 20^{12} = (2^2 \times 5)^{12} = 2^{24} \times 5^{12} \][/tex]

[tex]\[ = (2^4)^6 \times 5^{12} = 2^{24} \times 5^{12} \][/tex]

Now, we can rewrite the expression:

[tex]\[ \frac{15}{-(2^{24} \times 5^{12})} \times \frac{1}{p^4} \times \frac{1}{q^6} \times \frac{1}{q^{-3}} \][/tex]

[tex]\[ = -\frac{15}{2^{24} \times 5^{12} \times p^4 \times q^6 \times q^{-3}} \][/tex]

[tex]\[ = -\frac{15}{2^{24} \times 5^{12} \times p^4 \times q^3} \][/tex]

This is the simplified form of the expression.

Begin by using the cosine rule to find the unknown side.

Where a- unknown side, b is 12, c is 16 and angle in between is 108° 

The cosine rule...

        a² = b² + c² - 2bc cosA

            = 12² + 16² - (2×12×16 cos108°)

            = 518.655076

Taking the positive square root...

         a = 22.774 

         a = 22.8 cm nearest tenth

The question involves finding the length of the third side of a triangle using the Law of Cosines. The known parameters are two sides and the included angle. The formula c² = (12 cm)² + (16 cm)² - 2*(12 cm)*(16 cm)*cos(108 degrees) can be used to calculate the length of third side.

In this problem, you're dealing with a triangle and need to find the length of the third side. This is a classic case of using the Law of Cosines - a geometric rule that helps you find the length of a side of a triangle when you know the lengths of the other two sides and the angle in between (or in this case, SAS information).

The Law of Cosines formula is: c² = a² + b² - 2ab*cos(C), in which 'a' and 'b' represent the sides of the triangle adjacent to angle 'C', and 'c' is the other side. In your case, 'a' is 12.0 cm, 'b' is 16.0 cm and angle 'C' is 108.0 degrees. The third side 'c' can be calculated accordingly.

So, in this scenario, c² = (12 cm)² + (16 cm)² - 2*(12 cm)*(16 cm)*cos(108 degrees). With some basic arithmetic and the use of a calculator to compute the cosine, you can determine the length of the third side, 'c', to the nearest tenth of a cm.

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The dimensions of the rectangle are [tex]\( \boxed{9 \text{ km} \times 12 \text{ km}} \)[/tex].

Let's denote the length of the rectangle as [tex]\( l \)[/tex] kilometers, and its width as [tex]\( w \)[/tex] kilometers.

From the problem statement, we have two pieces of information:

1. The width is 6 kilometers less than twice the length:

  [tex]\[ w = 2l - 6 \][/tex]

2. The area of the rectangle is 108 square kilometers:

  [tex]\[ lw = 108 \][/tex]

Now we can substitute the expression for \( w \) from the first equation into the second equation:

[tex]\[l(2l - 6) = 108\][/tex]

Expand and simplify the equation:

[tex]\[2l^2 - 6l = 108\][/tex]

Subtract 108 from both sides to set the equation to zero:

[tex]\[2l^2 - 6l - 108 = 0\][/tex]

Divide every term by 2 to simplify:

[tex]\[l^2 - 3l - 54 = 0\][/tex]

Now, we'll solve this quadratic equation using the quadratic formula, [tex]\( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -54 \)[/tex]:

[tex]\[l = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1}\][/tex]

[tex]\[l = \frac{3 \pm \sqrt{9 + 216}}{2}\][/tex]

[tex]\[l = \frac{3 \pm \sqrt{225}}{2}\][/tex]

[tex]\[l = \frac{3 \pm 15}{2}\][/tex]

This gives us two possible solutions for [tex]\( l \)[/tex]:

[tex]\[l = \frac{18}{2} = 9 \quad \text{or} \quad l = \frac{-12}{2} = -6\][/tex]

Since length cannot be negative, we take [tex]\( l = 9 \)[/tex] kilometers.

Now, substitute [tex]\( l = 9 \)[/tex] back into the expression for [tex]\( w \)[/tex]:

[tex]\[w = 2l - 6 = 2 \cdot 9 - 6 = 18 - 6 = 12\][/tex]

Therefore, the dimensions of the rectangle are:

- Length [tex]\( l = 9 \)[/tex] kilometers

- Width [tex]\( w = 12 \)[/tex] kilometers

To verify, calculate the area:

[tex]\[l \times w = 9 \times 12 = 108 \text{ square kilometers}\][/tex]

Since this matches the given area, the dimensions [tex]\( l = 9 \)[/tex] kilometers and [tex]\( w = 12 \)[/tex] kilometers are correct.

Thus, the dimensions of the rectangle are [tex]\( \boxed{9 \text{ kilometers} \times 12 \text{ kilometers}} \)[/tex].

Answer is 8, i just did it

Answer:

B

Step-by-step explanation:

The approximate difference in the growth rate is 40 percent.

Answer:

a) monomial
b) constant
c) binomial
d) trinomial
e) binomial
f) constant
g) monomial

Step-by-step explanation:

Let's start by defining the terms in the question:
A monomial is an expression with a single term, regardless of how many variables are in the term.
A binomial is an expression with two different terms, which cannot be combined into a single term due to their variables not matching.
A trinomial is an expression with three different terms, which cannot be combined into a binomial or monomial because their variables do not match.
A constant is an integer that lacks a variable attached to it. An integer with a variable attached to it would be a monomial, and the integer itself would be called a coefficient.

Now, we can get into the parts of the question:
a) 8a²b is two variables however it is still a single term so this is a monomial.

b) -39 is an integer with no variables attached. Therefore, it is a constant.

c) x + y is two variables and two different terms that cannot be combined into one single term. It is a binomial.

d) -12xy + 5y - x² showcases three different variables and three different terms that cannot be consolidated. Thus, it is a trinomial.

e) 2x² + 2y² demonstrates two different variables and two different terms that are not alike. As such, we have a binomial.

f) 3/4 is an integer with no variables in sight. We have here a constant.

g) -32m²n² has two variables but is only a single term on its own. This is a monomial.

The area of the rectangle at L = 14 cm and B = 9 cm is increasing at 97 cm²/s.

The area of a rectangle is given by -

A[R] = L x B

Given is the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 5 cm/s.

Now, we can write -

dL/dt = 3 cm/s

dB/dt = 5 cm/s

We know, that the area is -

A = LB

differentiating both sides with respect to [t], we get -

dA/dt = L dB/dt + B dL/dt

dA/dt = 5L + 3B

At L = 14 cm and B = 9 cm.

(dA/dt) [14, 9] = 5 x 14 + 3 x 9 = 70 + 27 = 97 cm²/s

Therefore, the area of the rectangle at L = 14 cm and B = 9 cm is increasing at 97 cm²/s.

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Answer:

The slope of the function is 9.90

Step-by-step explanation:

You can find the slope by looking at the coefficient of the independent variable in the equation. In this case, we would be looking for the coefficient of n.

[tex] |\Omega|=52\\
|A|=26\\\\
P(A)=\dfrac{26}{52}=\dfrac{1}{2}=50\% [/tex]