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3 years ago

The question is in the picture.

Mathematics

1 answer:

bogdanovich [222]3 years ago

3 0

Just download photomath bro, does everything for you.

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4z−(−3z)=<br> combining like terms

uranmaximum [27]

For combining like terms: it is 7z

3 0

3 years ago

Read 2 more answers

Can anyone help? It’s not 14, I’ll give Brainly + 15 pts

Julli [10]

Short side length = x

Medium side = x + 7

Long side = 5x

Total 49, so

x + x + 7 + 5x = 49

If you combine like terms:

7x + 7 = 49

Subtract 7 from both sides

7x = 42

Divide both sides by 7

x = 6 inches.

8 0

3 years ago

"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0

3 years ago

Carman spent 42$ on 6 hats how much did each hat cost. Its for my sisters elementary school class and i dont even know the awnse

Travka [436]

Answer: 7

Step-by-step explanation:

Step 1: Since it is asking how much each hat costed. We would divide.

42$ divide 6 = 7$ per hat

8 0

3 years ago

A rectangular sign has length of 3 and 1/2 ft and a width of 1 and 1/8 ft. Will the area of the sign be greater or less than 3 a

zavuch27 [327]

Answer:

$Area = 3\dfrac{15}{16}\ sq\ ft$

Area is greater than $3\frac{1}{2}$ sq ft.

Step-by-step explanation:

Length of rectangular sign board = $3\frac{1}{2}$ ft

Width of rectangular sign board = $1\frac{1}{8}$ ft

Area of a rectangle is given by the formula:

$A = Length \times Width$

To find the area, we are going to multiply $3\frac{1}{2}$ with $1\frac{1}{8}$ .

$1\frac{1}{8}$ is greater than 1.

Therefore the multiplication will be greater than $3\frac{1}{2}$ .

Hence, we can say that area will be greater than $3\frac{1}{2}$ sq ft.

Area of the sign:

$A = 3\dfrac{1}{2}\times 1\dfrac{1}8\\\Rightarrow A = \dfrac{7}{2}\times \frac{9}8\\\Rightarrow A = \dfrac{63}{16}\\\Rightarrow A = 3\dfrac{15}{16}\ sq\ ft$

8 0

3 years ago