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Galina-37 [17]
2 years ago
14

6x^2 - 21 = 9 solve for x

Mathematics
1 answer:
earnstyle [38]2 years ago
8 0

Answer:

x= \sqrt{5} , -\sqrt{5}

Step-by-step explanation:

Move everything to left side and set it equal to 0

Solve using quadratic formula

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xxTIMURxx [149]

Answer:

Combine like terms

Step-by-step explanation:

The first step in such equations is to simplify the equation which is done most often by combining like terms, and by combining like terms i mean for example:

1 + 5 + 3x = 4y + z

You will combine the like terms (those that can be added or subtracted):

6 + 3x = 4y + z

In our example, combining terms would be adding 2x/3 and 1x/3 to give

3x/3 + 2 = 5

The equation is now simple and easy to solve as you simplify 3x/3 to 1x and then proceed to rearrange the equation to yield the value of x

Hope this helps!

6 0
2 years ago
Answer 5, 6 , and 7 by today please.
kherson [118]

Answer:

Step-by-step explanation:

5. a) ∠1 and ∠2 are remote interior angles of ∠ACD so that means that ∠ACD = ∠1 + ∠2

   b) Because an exterior angle is the sum of its two remote interior angles it makes sense that an exterior angle is greater in measure than either of its remote interior angles.

6. BD = DB  Reflexive property

    ∠3 = ∠5, ∠4 = ∠6  Alt. int. angles

    ΔADB = ΔCDB   ASA

7. AB = BC Def. of midpoint

   ∠1 = ∠2 Given

   ∠BAE = ∠CBD Corresponding angles

   ΔBAE = ΔCBD ASA

    ∠D = ∠E CPCTC

3 0
2 years ago
HELP it's Geometry HW I WILL GIVE BRAIN LIST AND 5 STARS​
Margaret [11]
Number one: if line segments SR & RT are perpendicular line segment Tu and US are perpendicular and angle STR is congruent to angle TSU, then triangle TRS is congruent to triangle SUT
Number two: if line segment AC is congruent to line segment CB and line segment CB bisects line segment AB, then < a is congruent to < B
3 0
3 years 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
A water wheel has a radius measuring between 13 feet and 24 feet. The wheel is able to turn 7π9 radians from its starting positi
madam [21]

Answer:

31.76 ft and 58.64 ft

Step-by-step explanation:

The radius measures between 13 feet and 24 feet.

The wheel is able to turn 7π/9 radians before getting stuck.

We need to find the range of distances that the wheel could spin before getting stuck. That is, the length of arc.

Length of an arc is given as:

L = \frac{\theta}{2\pi} * 2\pi r

where θ = central angle = 7π/9 radians

r = radius of the circle

Therefore, for 13 feet:

L = \frac{7\pi}{18 \pi} * 2 * \pi * 13\\\\L = 31.76 ft

For 24 feet:

L = \frac{7\pi}{18 \pi} * 2 * \pi * 24\\\\L = 58.64 ft

The wheel could spin between 31.76 ft and 58.64 ft before getting stuck.

4 0
3 years ago
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