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3 years ago

6

What is the function rule for the line? f(x)=−32x−2f(x)=−23x−2f(x)=32x−2f(x)=−32x+2A coordinate grid with x and y axis ranging f

rom negative five to five. The line passes the points begin ordered pair negative 2 comma 1 end ordered pair and begin ordered pair 0 comma negative 2 end ordered pair.

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

f(x)=23x−2

Explanation:

still trying to figure that out

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You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall

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Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are $\rightarrow 120ft5in+2(10ft)$

The two long are $\rightarrow 122ft1in=122.08ft$

The two shors are $\rightarrow 86ft4.5in = 86.375ft$

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

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Hence, we need order 5% plus that represent with the quantity

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A semiconductor is a solid substance that has a conductivity between that of an insulator and that of most metals. (True , False

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2 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

a)

$-V_{2} =2V_{1}$ ( double)

b)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

b)

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago