All categories

2 years ago

Regarding Linear questions

Mathematics

2 answers:

soldi70 [24.7K]2 years ago

8 0

See the attached pictures for the solutions:

viva [34]2 years ago

7 0

1.
2x + 10 = 30 - x
x = 20/3 = 6.67
y = 70/3 = 23.33333...
Both lines intersect at the 23.33... point

Blue Line: (0,10) , (20/3)(70/3)
Red Line: (0,30) , (12,17)

2.
Slope: (y2 - y1 ÷ x2 - x1)

= (-3 - 5) ÷ 3 - (-1)

= -6 ÷ 4

= -2

Slope = -2

Intercept:

y = mx + b

SLOPE = m

y = (-2x) + b

b = y1 + m

b = 5 + (-2) = 5 - 2 = 3

Intercept = 3

Hope This HELPS!

You might be interested in

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

How many minutes he does arm routine and abdominal routine

AlekseyPX

Let $x$ be the amount of time it takes to perform an arm routine and $y$ be the amount of time it takes to perform an abdominal routine. We see:

$4x+y=70$
$4x+2y=100$

Subtracting the second equation from the first gives $y=30$ . Substituting gives $4x+30=70$ , so $4x=40$ and $x=10$ .

Thus, an arm routine takes ten minutes and an abdominal routine takes thirty minutes.

5 0

3 years ago

A 270 feet escalator is leaning against a wall and makes a 53 angle with the floor. How high does the escalator reach the wall

r-ruslan [8.4K]

Answer:

x=215.63 ft

Step-by-step explanation:

sin(53）=x/270

x=270sin(53)

x=215.63 ft

5 0

3 years ago

Select from the drop-down menus to correctly complete the statement.

Dominik [7]

We are given the expression
5⁹ x 5³
We can solve this using the rule of exponents which is
a^n x a^m = a^(a + m)
Therefore, the answer is
to keep the base and add the exponents
5⁵⁺³ = 5⁸<span />

4 0

3 years ago

Read 2 more answers

The distance between two towns is 21.673 km. Round of the distance to nearest 0.01 km.

Volgvan

Answer:

21.67

Step-by-step explanation:

look at the number in the 0.01 value then the number next to it

if it's five or above then the 7 goes up if it's 4 and below then the number stays the same. in this case the number will stay the same

6 0

2 years ago