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DENIUS [597]
1 year ago
14

Length of the base of the given triangle is (x+3) units and the perpendicular height is (x-1)units. If the area of the triangle

is 10 square units, show that x satisfy the equation x2 + 2x - 23 = 0 Solve above equation and find the value of x to the nearest whole number and calculate the perpendicular height of the triangle. (Take V6 = 2.45)​
Mathematics
1 answer:
Lana71 [14]1 year ago
7 0

Step-by-step explanation:

the area of a triangle is

baseline × height / 2

so, in our case we know this is 10 :

(x+3)(x-1)/2 = 10

(x+3)(x-1) = 20

x² - x + 3x - 3 = 20

x² + 2x - 23 = 0

that is exactly the provided equation. so, yes, x satisfies it.

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -23

x = (-2 ± sqrt(2² - 4×1×-23))/(2×1) =

= (-2 ± sqrt(4 + 92))/2 = (-2 ± sqrt(96))/2 =

= (-2 ± sqrt(16×6))/2 = (-2 ± 4×sqrt(6))/2 =

= -1 ± 2×sqrt(6)

x1 = -1 + 2×sqrt(6) = 3.898979486... ≈ 4

or using sqrt(6) ≈ 2.45

= -1 + 2×2.45 = -1 + 4.9 = 3.9 ≈ 4

x2 = -1 - 2×sqrt(6) = -5.898979486... ≈ -6

or using sqrt(6) ≈ 2.45

= -1 - 2×2.45 = -1 - 4.9 = -5.9 ≈ -6

x2 would give us negative lengths for the triangle, which does not make sense.

so, x = 4 is our solution.

and that makes the height x-1 = 4-1 = 3 units.

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-4x+3x=2 how do I solve for x?​
Snezhnost [94]

Answer: Combine like terms!

Step-by-step explanation:

-4x + 3x = 2

-1x = 2

-1x/-1 = 2/-1

x = -2

7 0
3 years ago
Convert the following units of time. a. 5 years to days b. 8 weeks to days c. 7 days to weeks d. 336 hours to seconds e. 12 hour
AysviL [449]

Answer:

A. 1,825 days

B. 56 days

C. 1 week

D.  1,209,600 seconds

E. 43,200 seconds

F. 2 hours

Step-by-step explanation:

A. 1 year = 365 days, then

5 years = 5\cdot 365 days = 1,825 days.

B. 1 week = 7 days, then

8 weeks = 8\cdot 7 days = 56 days.

C. 7 days = 1 week.

D. 1 hour = 3,600 seconds, then

336 hours = 336\cdot 3,600 seconds = 1,209,600 seconds.

E 12 hours = 12\cdot 3,600 seconds = 43,200 seconds.

F. 3,600 seconds = 1 hour, then

7,200 seconds = \dfrac{7,200}{3,600} hours = 2 hours.

3 0
3 years ago
Read 2 more answers
How do i write factored form for the graphs?
xxMikexx [17]
1. Take the two x-axis intersections.

For Q 31 it is:
x = -4 and x = 0

2. Reorganise the previous equations

x +4 = 0 and x = 0

3. Put the x side in brackets

x(x+4)

4. Done

—-

Q 32

x = -1
x = 5

x + 1 = 0
x - 5 = 0

(x+1)(x-5)
5 0
2 years ago
Read 2 more answers
N Identify the next (3) numbers in the set: 6, 36, 2, 4, -30... *
sveticcg [70]

Answer:

-60, -120, -154

Step-by-step explanation:

-60, -120, -154

because

from 6 to get to 36, it is multiplied by itself.

from 36 to get to 2, it is subtracted by 34.

the pattern continues.

Can I get brainliest please? Thanks! Hope this helps!

8 0
3 years ago
Read 2 more answers
An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click
Sergeeva-Olga [200]

Answer:

A.    \mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

B.    \mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

Step-by-step explanation:

Given that:

An investment of  Amount = $8000

earns  at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A)  Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

A = Pe^{rt}

where;

A = (8000) \ e ^{0.7t}

The instantaneous rate of change = \dfrac{dA}{dt}

\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )

= 8000 \dfrac{d}{dt}e^{0.07 \ t}

\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 2 years; the instantaneous rate of change is:

\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}

\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

12000 = (8000)e^{0.07 \ t}

\dfrac{12000 }{8000}= e^{0.07 \ t}

1.5= e^{0.07 \ t}

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

\dfrac{dA}{dt}= 560 e^{0.07 \ t}

At t = 5.79

\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}

\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }

7 0
3 years ago
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