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uysha [10]
3 years ago
8

Solve for c c+d=15 3c+2d=36 WILL MARK BRAINLIEST AND MANY POINTS PLS ANSWER

Mathematics
2 answers:
alexira [117]3 years ago
7 0
C = 6

c = 15 - d ( get c by itself first)

3(15 - d) + 2d = 36 (replace c with 15 + d)
45 - 3d + 2d = 36 (distribute)
45 - d = 36 (combine like terms)
-d = -9
d = 9
klio [65]3 years ago
5 0
C is 6 & d is 9

6+9=15
3(6)+2(9)=36
18 + 18 = 37
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Can someone please explain how to go the answers for this
Effectus [21]

Answer:

A=3, B=4, C=-5

Step-by-step explanation:

So A represents the amplitude, which is the distance between the midline (C), the middle of the graph, and the maximum (or minimum). B is the distance of the period of the graph. In this case, we see that the midline is C=-5 because that is the halfway point of the sine function. We also see that the period distance is B=5-1=4, so our period would be 2π/4 or π/2. Our amplitude would be A=3 because |a|=|-5-(-2)|=|-5+2|=|-3|=3. Observe the graph to see this visually.

7 0
3 years ago
Ken can walk
pentagon [3]

Answer:

Hey.  I'm not sure if you have typos, but if you're saying ''Ken can walk 40 dogs in 8 hours, how many dogs can Ken walk in 12 hours", the answer is 60.

Step-by-step explanation:

So, we get our answer by dividing the amount of dogs Ken can walk by the amount of hours.  40/8 = 5.  He walks 5 dogs an hour.  Now that we know he walks 5 dogs an hour, we can multiply by 5 for each hour.  5 (Dogs/Hour) x 12 (Hours the dogs were walked) = 60, and that is our final answer.

7 0
3 years ago
Juana tiene en su cuenta bancaria $98.857. Si expide un cheque por $39.948 y deposita otro por $10.845 ¿Cuánto dinero le queda a
Sonja [21]

Answer:

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Step-by-step explanation:

5 0
3 years ago
Una diseñadora de modas que confecciona trajes de noche, tarda 3 horas en cortar y 2 horas en coser cada vestido de fiesta. Para
ella [17]

For this case we define the following variables:

x: Number of party dresses

y: Number of suits

You have 30 hours per week to cut, that is, the first equation is given by:

3x + 3y = 30\\

It is also known that 25 hours per week are available for sewing, that is:

2x + 3y = 25\\

It has a system of two equations with two unknowns, solving we have:

3x + 3y = 30\\\\2x + 3y = 25\\

Multiplying the second equation by -1:

3x + 3y = 30\\\\-2x-3y = -25\\

Adding up:

x + 0 = 5\\\\x = 5\\

Substituting x in the first equation:

3 * 5 + 3y = 30\\\\15 + 3y = 30\\

Clearing and:

3y = 30-15\\\\3y = 15\\

y = \frac{15}{3}\\\\y = 5\\

Thus, per week, the designer can produce 5 party dresses and 5 suits working at her maximum capacity.

Answer:

5 Party dresses

5 Suits


5 0
3 years ago
Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit
dem82 [27]

Answer:

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

7 0
3 years ago
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