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S_A_V [24]
3 years ago
10

Please help ASAP ill appreciate it a lot

Mathematics
1 answer:
trapecia [35]3 years ago
6 0

the correct answer is 5/12

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The ratio of a to b is 3 to 1, and the ratio of b to c is 1 to 5. What is the value of 2a+3b/ 4b+3c?
Nostrana [21]

Answer:

B

Step-by-step explanation:

Expressing the ratios as fractions, that is

\frac{a}{b} = \frac{3}{1} ( cross- multiply )

a = 3b

\frac{b}{c} = \frac{1}{5} ( cross- multiply )

c = 5b

------------------------------------------

Given

\frac{2a+3b}{4b+3c} , substitute values from above for a and c

= \frac{2(3b)+3b}{4b+3(5b)}

= \frac{6b+3b}{4b+15b}

= \frac{9b}{19b} ← cancel b on numerator/ denominator

= \frac{9}{19} → B

6 0
3 years ago
Which of the following finance options will have the lowest total financed
Pavlova-9 [17]

Answer:

A is the answer

Step-by-step explanation:

did the test hope this helps

8 0
2 years ago
A person on a moving sidewalk travels 6 feet in 3 seconds. The moving sidewalk has a length of 80 feet. How long will it take to
tekilochka [14]

Answer:

40.5 seconds

Step-by-step explanation:

5 0
1 year ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
2 years ago
Pls help me !!!!!!!!!!!!!!!!!!!!!!!!
Vikentia [17]

Answer:

9.89 m

Step-by-step explanation:

AC is the hypotenuse (H)

x is the opposite (O)

O and H are in sine

sin=O/H

H=O/sin

H=7.9/sin(53)

=9.89187169943418

3 SF

9.89 m

8 0
3 years ago
Read 2 more answers
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