<img src="https://tex.z-dn.net/?f=18%20%7Bx%7D%5E%7B2%7D%20-%20x%20-%2039%20" id="TexFormula1" title="18 {x}^{2} - x

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bekas [8.4K]

3 years ago

"18 {x}^{2} - x - 39 " align="absmiddle" class="latex-formula">
How to solve this

Mathematics

2 answers:

Anton [14]3 years ago

7 0

I don’t know if you can solve it since it is an expression?

Maru [420]3 years ago

7 0

The answer is X=1.5

x squared = 2.25 (1.5 x 1.5)

18 x 2.25 = 40.5

40.5 - 1.5 = 39

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To overcome an infection an anti-biotic is injected into David’s bloodstream. After the injection the anti-biotic in his body de

meriva

Answer:

4.45 ccs will remain after 5 hours.

Step-by-step explanation:

The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.

(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)

(dQ/dt) = -kQ

(dQ/Q) = -kdt

∫ (dQ/Q) = -k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.

We obtain

In (Q/Q₀) = -kt

(Q/Q₀) = e⁻ᵏᵗ

Q(t) = Q₀ e⁻ᵏᵗ

the initial injection was 8 ccs and 5 ccs remain after 4 hours

Q₀ = 8 ccs,

At t = 4 hours, Q = 5 ccs

5 = 8 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.625

-kt = In (0.625) = -0.47

-4k = 0.47

k = 0.1175 /hour

Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ

At t = 5 hours, Q = ?

Q = 8 e⁻⁰•¹¹⁷⁵ᵗ

0.1175 × 5 = 0.5875

Q = 8 e(^-0.5875)

Q = 4.45 ccs

Hope this Helps!!!

8 0

3 years ago

Dy If y - 2x+3 then 3x+29 dx

Sergeu [11.5K]

Answer:

$y'=\frac{-5}{(3x+2)^2}$

Step-by-step explanation:

Step 1: Write equation

$y=\frac{2x+3}{3x+2}$

Step 2: Find derivative

Quotient Rule: $y'=\frac{(3x+2)(2)-(3)(2x+3)}{(3x+2)^2}$
Simplify: $y'=\frac{(6x+4)-(6x+9)}{(3x+2)^2}$
Simplify: $y'=\frac{-5}{(3x+2)^2}$

6 0

3 years ago

Gina started with the circle (x−2)2+y2=16. She translated it 4 units left and 1 unit down. Then she reduced the radius by 1 unit

olchik [2.2K]

Answer:

(x + 2)² + (y + 1)² = 9

Step-by-step explanation:

Gina started with a circle with the given equation as,

(x - 2)² + (y - 0)² = 16

Since, standard equation of a circle is,

(x - a)² + (y - b)² = r²

Here (a, b) is the center of the circle and 'r' is the radius.

Therefore, from the given equation of the circle,

Center of the circle → (2, 0)

Radius of the circle → √16 = 4 units

She translated this circle 4 units left and 1 unit down.

Rule for the new center after translation will be,

(x, y) → (x - 4, y - 1)

(2, 0) → (2 - 4, 0 - 1)

→ (-2, -1)

Coordinates of the new center → (-2, -1)

Then she reduced the radius by 1 units.

Therefore, radius of the new circle = 4 - 1 = 3 units

Now we substitute these values in the standard equation of the circle to get the equation of the new circle.

(x + 2)² + (y + 1)² = 3²

(x + 2)² + (y + 1)² = 9

5 0

3 years ago

Quick help please it will mean a lot if you help me :)

Ganezh [65]

Answer:

Green heart is 2 and blue shape is 8.

Step-by-step explanation:

We know 1 purple drop equals 3.

So 3+3 = 6

The total weight is 16 so we can do 16-6 = 10

We have 10 left. Since we need the weight to be balanced the green heart must be 2.

Because then it would be 8 on 1 side (2 purples equals 6 and 1 heart equals 2, so 6+2=8) and 8 on the other side (8+8=16).

8 0

3 years ago

Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily,

Rudik [331]

Answer:

Part 1) 8.17 years

Part 2) 4.98 years

Part 3) 4.95 years

Part 4) 4.95 years

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part 1) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded annually

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=1$

substitute in the formula above

$2p=p(1+\frac{0.14}{1})^{t}$

$2=(1.14)^{t}$

Apply log both sides

$log(2)=log[(1.14)^{t}]$

$log(2)=(t)log(1.14)$

$t=log(2)/log(1.14)$

$t=8.17\ years$

Part 2) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded monthly

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=12$

substitute in the formula above

$2p=p(1+\frac{0.14}{12})^{12t}$

$2=(\frac{12.14}{12})^{12t}$

Apply log both sides

$log(2)=log[(\frac{12.14}{12})^{12t}]$

$log(2)=(12t)log(\frac{12.14}{12})$

$t=log(2)/12log(\frac{12.14}{12})$

$t=4.98\ years$

Part 3) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded daily

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=365$

substitute in the formula above

$2p=p(1+\frac{0.14}{365})^{365t}$

$2=(\frac{365.14}{365})^{365t}$

Apply log both sides

$log(2)=log[(\frac{365.14}{365})^{365t}]$

$log(2)=(365t)log(\frac{365.14}{365})$

$t=log(2)/365log(\frac{365.14}{365})$

$t=4.95\ years$

Part 4) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% continuously

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14$

substitute in the formula above

$2p=p(e)^{0.14t}$

Simplify

$2=(e)^{0.14t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.14t}]$

$ln(2)=(0.14t)ln(e)$

Remember that ln(e)=1

$ln(2)=(0.14t)$

$t=ln(2)/(0.14)$

$t=4.95\ years$

4 0

3 years ago