Which equations represent the line that is parallel to 3x − 4y = 7 and passes through the point (−4, −2)? Check all that apply.

A. y = –x + 1
B. 3x − 4y = −4
C. 4x − 3y = −3
D. y – 2 = –(x – 4)
E. y + 2 = (x + 4)

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

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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Answer:

B

Step-by-step explanation:

The approximate difference in the growth rate is 40 percent.

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.

To add 8 3/10 and 7 2/15, we need to find a common denominator. The sum is 15 1/15.

To add 8 3/10 and 7 2/15, we need to find a common denominator. The least common multiple of 10 and 15 is 30. So we convert both fractions to have a denominator of 30: 8 3/10 = 8 9/30 and 7 2/15 = 7 4/30. Now we can add the whole numbers and the fractions separately:

Combining the results, the sum is 15 13/30. To express the answer in lowest terms, we can simplify the fraction: 13 and 30 are both divisible by 13, so dividing both by 13 gives us 1 and 2/30. Therefore, the final answer is 15 2/30, which can be further simplified to 15 1/15.

Answer:

Options A, B and E represent mutually exclusive events.

Step-by-step explanation:

Two events are mutually exclusive that can not happen at a same time.

Now let us look at our options and find which statements represent mutually exclusive events.

A. We can see that landing on an unshaded portion and landing on 2 are mutually exclusive events as unshaded portion contains 3 and 4 and 2 lies on shaded portion.

B. Landing on a shaded portion and landing on 3 is also mutually exclusive as 3 lies on unshaded portion.

C. Landing on a shaded portion and landing on an even number can happen at the same time as 2 is an even number and it lies on shaded portion.

D. Landing on an unshaded portion and landing on a number greater than 3 can happen at the same time as 4 in greater than 3 and it lies on unshaded portion.

E. Landing on a shaded portion and landing on an unshaded portion  are mutually exclusive events as spinner can be either in shaded portion or in unshaded portion.

Therefore, Options A, B and E are our correct choices.

Answer:

A, B, E

Step-by-step explanation:

I just did it on edge and got it right

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:

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.

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.

Answer is 8, i just did it

The width of the construction paper is 12 inches.

The formula for area is length(l)*width(w)=area in units squared(a^2).

To find the width, we divide both sides by length, this gives us the following:

17*w=204in^2     w=204in^2/17  

To find the answer, you just perform the operation.

204/17=12

The width is 12 inches.