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2 years ago

Can you simplify 4(-5+100)

1 answer:

8 0

380
Do 4(-5) = -20
Then 4(100) = 400
Add -20+400=380
(You have to subtract because of the negative number but the answer is positive because the bigger number in the equation is positive)

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What’s the answer please help

sp2606 [1]

Answer:

{121,144,169} X intersecting Y means common number between x and y set

5 0

3 years ago

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

4 years ago

What is Question I (if m<EBF = 117°, fine m<ABE)

Natasha_Volkova [10]

Answer:

m∠ABE = 27°

Step-by-step explanation:

* Lets look to the figure to solve the problem

- AC is a line

- Ray BF intersects the line AC at B

- Ray BF ⊥ line AC

∴ ∠ABF and ∠CBF are right angles

∴ m∠ABF = m∠CBF = 90°

- Rays BE and BD intersect the line AC at B

∵ m∠ABE = m∠DBE ⇒ have same symbol on the figure

∴ BE is the bisector of angle ABD

∵ m∠EBF = 117°

∵ m∠EBF = m∠ABE + m∠ABF

∵ m∠ABF = 90°

∴ 117° = m∠ABE + 90°

- Subtract 90 from both sides

∴ m∠ABE = 27°

4 0

4 years ago

Read 2 more answers

Help? what should i choose

igomit [66]

Answer:

Step-by-step explanation:

Angles on a straight line add up to 180 so:

180-110 = 70

As this is an isosceles triangle,

y must be 70.

We know y is 70 and there are 2y and one x. A triangle also adds up to 180.

This means:

2y+x = 180

2(70)+x = 180

140+x = 180

x = 40

x = 40 y = 70

8 0

3 years ago

What are the values of x and y?<br><br> (Explain please it will help a lot,thanks)

Kitty [74]

X = 80
y = 85

If you look at only the left triangle, you can see that x is part of one of the angles. The 3 interior angles of the triangle add up to 180 degrees, so subtracting the other two angles, you get x + 30 = 110, or x = 80.

The same method can be used for the other triangle to find y.

5 0

4 years ago