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2 years ago

How many solutions exist for the system of equations graphed below?

Mathematics

1 answer:

Lapatulllka [165]2 years ago

4 0

Answer: one

Step-by-step explanation:

When a system of equations is graphed the solutions are the point of intersections. Here, we only have one point of intersection since these are both linear equations. That means there is only one solution.

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Solve for q<br><br>8q-15=49

Molodets [167]

Answer:

q=8

Step-by-step explanation:

To solve for q, we need to get q by itself. To do this, preform the opposite of what is being done to the equation. Also, everything must be done to both sides of the equation.

8q-15=49

Add 15 to both sides, since 15 is being subtracted.

8q-15+15=49+15

8q=64

Divide both sides by 8, since 8 and x are being multiplied.

8q/8=64/8

q=8

8 0

3 years ago

Read 2 more answers

The function F is defined by F(x)= 12/x + 1/2 . Use this formula to find the following values of the function.

IgorLugansk [536]

Answer:

$F(3)=\frac{9}{2} \\F(-12)=\frac{-1}{2} \\F(\frac{1}{3} )=\frac{73}{2} \\F(\frac{3}{4} )=\frac{33}{2}$

Step-by-step explanation:

Given: $F(x)= \frac{12}{x} + \frac{1}{2}$

To find: Values of the function F(3) , F(−12) , $F(\frac{1}{3})$ , $F(\frac{3}{4})$

Solution:

A function is a relation in which each and every element of the domain has a unique image in the co-domain.

To find the values of the functions $F(3),F(-12),F(\frac{1}{3} ),F(\frac{3}{4} )$ , put $x=3, -12, \frac{1}{3},\frac{3}{4}$ in the given function $F(x)= \frac{12}{x} + \frac{1}{2}$

$F(x)= \frac{12}{x} + \frac{1}{2}\\F(3)= \frac{12}{3} + \frac{1}{2}\\=4 + \frac{1}{2}\\=\frac{9}{2}$

$F(x)= \frac{12}{x} + \frac{1}{2}\\F(-12)= \frac{12}{-12} + \frac{1}{2}\\=-1+\frac{1}{2}\\ =\frac{-1}{2}$

$F(x)= \frac{12}{x} + \frac{1}{2}\\\\F(\frac{1}{3} )= \frac{12}{\frac{1}{3}} + \frac{1}{2}\\=36+\frac{1}{2}\\ =\frac{73}{2}$

$F(x)= \frac{12}{x} + \frac{1}{2}\\\\\\F(\frac{3}{4} )= \frac{12}{\frac{3}{4}} + \frac{1}{2}\\\\=16+\frac{1}{2}\\ \\=\frac{33}{2}$

8 0

3 years ago

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62.4% of what number is 17.16

e-lub [12.9K]

$\frac{62,4}{100}*x=\frac{1716}{100}\\
62,4*x=1716\\
x=27,5
$

4 0

3 years ago

Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the

ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let $r$ be the radius, $d$ the diameter, and $V$ the volume of the spherical balloon.

We know $\frac{dV}{dt}=20 \:{in^3/s}$ and we want to find $\frac{dr}{dt}$

The volume of a spherical balloon is given by

$V=\frac{4}{3} \pi r^3$

Taking the derivative with respect of time of both sides gives

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

We now substitute the values we know and we solve for $\frac{dr}{dt}$ :

$d=2r\\\\r=\frac{d}{2}$

$r=\frac{12}{2}=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044$

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and $\frac{dr}{dt}$ is:

$\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025$

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

8 0

3 years ago

If A = (0,0) and B = (2,5), what is the approximate length of AB?

Bad White [126]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{(2-0)^2+(5-0)^2}\implies AB=\sqrt{2^2+5^2} \\\\\\ AB=\sqrt{29}\implies AB\approx 5.39$

8 0

3 years ago

Read 2 more answers