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3 years ago

Solve the system of linear equations by elimination. 2x – y=0 3x – 2y = -3

Mathematics

1 answer:

devlian [24]3 years ago

6 0

Answer:

Point form: (3,6)

Equation form: x=3, y=6

Step-by-step explanation:

Add the equations in order to solve for the first variable. Then plug this value into the other equations in order to solve for the remaining variables.

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A truck driver drove 48 miles for 45 minutes. At this rate, how many miles can the truck driver drive in one hour?

alexandr1967 [171]

Answers

2880

Step-by-step explanation:

48 miles = 1 minute

? = 60 minutes( 1hr )

1hr = 48 × 60 = 2880miles

7 0

2 years ago

PLLSS HELP ME ty :)))

ArbitrLikvidat [17]

I think its 210 ≤ t

6 0

2 years ago

Consider the following pair of equations:

V125BC [204]

The pair of equations:
y = x + 4
y = -2x - 2

To solve the system of equation by substitution, you would substitute one variable for it's equivalent expression. Combining the pair of equations into one equation in only one variable. The easiest way to do this is by substituting out the variable y because it is already alone on one side of the equal sign. Substitute the y in the second equation for: x+4

x+4 = -2x - 2
solve for x
x + 2x = -2 -4
3x = -6
x = -2

now use the value for x to find y by putting into either equation.

y = x + 4
y = (-2) + 4
y = 2

solution is: (-2, 2)

4 0

2 years ago

Given that f(x) = X + 3 2 a) Find f(2) b) Find f'(x) c) Find f"(12)

joja [24]

Answer:

Step-by-step explanation:

Unclear. Did you mean f(x) = 2^(x + 3)? If so, the parentheses are mandatory.

Assuming that f(x) = 2^(x + 3) is correct, then:

a) f(2) = 2^(2 + 3) = 2^5 = 32

b) f'(x) = 2^(x + 3)*(x + 3)' = 2^(x + 3)(1) = 2^(x + 3)

c) f"(x) = 2^(x + 3) (same as in (b)), so that:

f"(12) = 2^15

8 0

2 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0

2 years ago