If −15 + 3x = −7 + 5x, then 6x =_____?

Answer:

(0,-5)

Step-by-step explanation:

move the x point to the right three times im pretty sure

Graph the linear equation. 3x-4y= 24

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

Let x = 0, then y = -6

Let y = 0, then x = 8

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

Plot (0,-6) and (8,0) and draw a line thru them to get:

Answer:

See below

Step-by-step explanation:

Given the expression

x(x-3) = 0

This can be written as x = 0 and x - 3 = 0

x = 0 and x = 0+3

x = 0 and x = 3

Hence the equation has 2 solutions since it is a quadratic equation

B) Given x^2 = -81

Take the square root of both sides

√x² = ±√-81

x =±√-81

Note that √-1 = i

x = ±9i

x = 9i and -9i

Hence this equation has 2 solutions because of its quadratic nature

C) (x−1)(x−1)=0

x - 1 = 0 and x = 1 = 0

x = 1 twice

Hence this equation has 1 solution because of its repitition of root

Answer:

[tex](a)\ -16t^2 + 7t + 63=0[/tex]

[tex](b)\ t = 2.215[/tex]

Step-by-step explanation:

Given

[tex]h(t) = -16t^2 + 7t + 63[/tex]

Solving (a): Equation when it hits the ground.

This means that [tex]h(t) = 0[/tex]

So, we have:

[tex]h(t) = -16t^2 + 7t + 63[/tex]

[tex]-16t^2 + 7t + 63=0[/tex]

Solving (b): The value of t in (a)

[tex]-16t^2 + 7t + 63=0[/tex]

Using quadratic formula, we have:

[tex]t = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]

This gives:

[tex]t = \frac{-7 \± \sqrt{7^2 - 4*-16*63}}{2*-16}[/tex]

[tex]t = \frac{-7 \± \sqrt{49+ 4032}}{2*-16}[/tex]

[tex]t = \frac{-7 \± \sqrt{4081}}{-32}[/tex]

[tex]t = \frac{-7 \± 63.88}{-32}[/tex]

Split

[tex]t = \frac{-7 + 63.88}{-32}; or\ t = \frac{-7 - 63.88}{-32}[/tex]

[tex]t = \frac{56.88}{-32}; or\ t = \frac{-70.88}{-32}[/tex]

[tex]t = -1.7775; or\ t = 2.215[/tex]

Time can't be negative; So:

[tex]t = 2.215[/tex]

Answer:

The exact length of AB is [tex]3 + 4\sqrt{10}[/tex] milimeters.

Step-by-step explanation:

Both triangles ACD and BCD are isosceles and triangles AEC, ADE, BDE and BCE are right-angled, where E is the point where line segments AB and CD meet each other. We can determine the exact length of AB by means of two horizontal right triangles (i.e. AEC, BCE) and the Pythagorean Theorem:

[tex]AB = \sqrt{AC^{2}-CE^{2}}+\sqrt{CB^{2}-CE^{2}}[/tex]

If we know that [tex]AD = AC[/tex], [tex]BC = BD[/tex], [tex]AC = 5\,mm[/tex], [tex]BC = 12\,mm[/tex] and [tex]CE = 4\,mm[/tex], then the exact length of AB is:

[tex]AB = \sqrt{(5\,mm)^{2}-(4\,mm)^{2}}+\sqrt{(12\,mm)^{2}-(4\,mm)^{2}}[/tex]

[tex]AB = 3 + 4\sqrt{10}\,[mm][/tex]

So, Vanessa's score = v. Let's work from there.

We know that 59 is decreased by Vanessa's score, which is subtraction, and thus: 59 - v

Then, we're told that that quantity is 7. "Is" can be used in place of an equals sign: = 7

Therefore, our equation is as follows: 59 - v = 7

Hope this helps!! :)

Answer:

See Explanation

Step-by-step explanation:

Given

[tex]b(t) = 3000 * 2^t[/tex]

The question is incomplete;

A possible question could be to calculate the initial amount of beetle

An exponential function is represented as:

[tex]f(x) = ab^x[/tex]

Where

[tex]a \to Initial\ amount[/tex]

By comparison:

[tex]a = 3000[/tex] --- The initial number of beetles

Another question could be to solve for f(t), after a certain number of weeks (say 3)

Substitute 3 for t in: [tex]b(t) = 3000 * 2^t[/tex]

[tex]b(3) = 3000 * 2^3[/tex]

[tex]b(3) = 3000 * 8[/tex]

[tex]b(3) = 24000[/tex]

9514 1404 393

Answer:

  C  log3(√((x -4)/x)

Step-by-step explanation:

The applicable rules of logarithms are ...

  log(a/b) = log(a) -log(b)

  log(a^n) = n·log(a)

The base is irrelevant, as long as all logs are to the same base.

__

  [tex]\dfrac{1}{2}(\log_3{(x-4)}-\log_3{(x)})=\dfrac{1}{2}\log_3(\dfrac{x-4}{x})=\boxed{\log_3\left(\sqrt{\dfrac{x-4}{x}}\right)}[/tex]

Answer:

0 ≤ x ≤ 30

Step-by-step explanation:

In the month of September, a bird population increased by a factor of 1.25 every day for those 30 days. The function below shows the number of birds, f(x), after x days:

f(x) = 250(1.25)ˣ

Which of the following is a reasonable domain for the function?

0 ≤ x ≤ 250

0 ≤ x ≤ 30

All positive integers greater than 100

All real numbers

Solution:

Given that f(x) = 250(1.25)ˣ, where x is the number of days and f(x) is the population of the birds after x days. This means that at the beginning of the month of September (i.e. at x = 0), the population of the birds was 250. And as each day goes by, the population of the bird increases by a factor of 1.25.

The domain of a function is the set of all possible independent variables. In this case the number of days (x) is the independent variable. Since in the month of September there are 30 days, therefore the domain is:

0 ≤ x ≤ 30

Answer:

4x-6=30

4x=36

x=9

hope this helps

have a good day :)

Step-by-step explanation:

Answer:

x = 9

Explanation:

Move all terms that do not contain x to the right side and solve.

Answer: x = 2

Step-by-step explanation:

4x + 4 = 2x + 8

2x = 4

x = 2

Answer: x = 2.

Step-by-step explanation: 4x+4=2x+8 subtract 4 from both sides.

4x = 2x+4 subtract 2x on both sides

2x = 4        divide both sides by two

x = 2

I hope this helps, have a nice day.

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]y^\frac{2}{3}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

The rule for a fraction exponent is:

[tex]a^{\frac{x}{y}}=\sqrt[y]{a^x}\\\\\\\huge\boxed{\frac{\text{Power}}{\text{Root}}}[/tex]

⸻⸻⸻⸻

We are given the expression of [tex]\sqrt[3]{y^2}[/tex].

⸻⸻⸻⸻

[tex]\boxed{\text{Rewriting the expression:}}\\\\\sqrt[3]{y^2}\\\\\text{The '3' is the root and the '2' is the power.}\\\\\rightarrow \sqrt[3]{y^2} \\\\\rightarrow {y^{\frac{2}{3}}[/tex]

⸻⸻⸻⸻

[tex]\text{The answer should be: } \boxed{y^\frac{2}{3} }[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

Increase in her mass = 3.472 kg

Her new mass = 46.872 kg

Step-by-step explanation:

Her original mass = 43.4 kg

Percentage increase in her mass = 8%

Increase in her mass = 8% of 43.4 kg

= 8/100 * 43.4 kg

= 0.08 * 43.4 kg

= 3.472 kg

Increase in her mass = 3.472 kg

b) Work out her new mass

Her new mass = Her original mass + Increase in her mass

= 43.4 kg + 3.472 kg

= 46.872 kg

Her new mass = 46.872 kg

Answer:

[tex]-4[/tex]

Step-by-step explanation:

Step 1: Apply the rule

[tex]\left(-64\right)=-64\\\\=\sqrt[3]{-64}[/tex]

Step 2: Apply the radical rule

[tex]\sqrt[3]{-64}=-\sqrt[3]{64}\\\\=-\sqrt[3]{64}[/tex]

Step 3: Factor the number

[tex]\:64=4^3\\\\=-\sqrt[3]{4^3}[/tex]

Step 4: Apply the radical rule

[tex]\sqrt[3]{4^3}=4\\\\=-4[/tex]

Therefore, [tex]\:\sqrt[3]{\left(\:-\:64\right)}\:=\:-4[/tex]

Answer:

-4

Step-by-step explanation:

Answer:

mp=160

dis=40

now,

sp=mp-dis

=Rs. 160-Rs.40

=Rs. 120

Answer:

996

Step-by-step explanation:

largest 3 digit number is 999

then just reduce slowly to find ans

Answer:

The largest 3-digit number divisible by 6 is 996.

Answer:

no  x = -8/13

Step-by-step explanation:

6x – (3x + 8) = 16x

Combine like terms

6x -3x -8 = 16

3x-8 = 16

Subtract 3x from each side

3x-8-3x = 16-3x

-8 = 13x

Divide by 13

-8/13 = 13x/13

-8/13 =x

Answer:

No, the answer would be -8/13

Step-by-step explanation:

6x-(3x+8)=16x

6x-3x-8=16x

3x-16x=8

-13x=8

x=-8/13

Answer:

9 3/8 inches

Step-by-step explanation:

Inches of candle : hours it burned

= 5 inches : 4 hours

How far has it burned after 7 1/2 hours?

Let

x = how far it has burned (inches)

Inches of candle : hours it burned = x inches : 7 1/2 hours

Equate both ratios

5 inches : 4 hours = x inches : 7 1/2 hours

5/4 = x ÷ 7 1/2

5/4 = x ÷ 15/2

5/4 = x * 2/15

5/4 = 2x/15

Cross product

5 * 15 = 4 * 2x

75 = 8x

x = 75/8

x = 9 3/8 inches

x = how far it has burned (inches) = 9 3/8 inches

Answer:

86.70 cm^2

Step-by-step explanation:

circumference = 11 x 3 = 33

circumference of a circle =  nD

33 = 3.14 X D

D = 10.51

r = 10.51 / 2 = 5.25

Area = nr^2

3.14 x 5.25^2 = 86.70 cm^2

Answer:

36 feet

Step-by-step explanation:

Given

[tex]h(t) = -16t^2 + 48t[/tex]

Required

Determine the maximum height attained

First, calculate the time to reach maximum height;

In a quadratic equation;

[tex]y = ax^2 + bx + c[/tex]

The maximum is:

[tex]x = -\frac{b}{2a}[/tex]

So, we have:

[tex]t = -\frac{b}{2a}[/tex]

Where

[tex]a = -16; b = 48[/tex]

So:

[tex]t = -\frac{b}{2a}[/tex]

[tex]t = -\frac{48}{2 * -16}[/tex]

[tex]t = -\frac{48}{-32}[/tex]

[tex]t = \frac{48}{32}[/tex]

[tex]t = 1.5[/tex]

The maximum height is at: [tex]t = 1.5[/tex]

So, we have:

[tex]h(t) = -16t^2 + 48t[/tex]

[tex]h_{max}=-16 * 1.5^2 + 48 * 1.5[/tex]

[tex]h_{max}=36[/tex]

Answer:

hmm very diffecult.

Step-by-step explanation:

if f (x) = StartRoot

Hi there! Assume that this is your question.

[tex] \large{ \int \limits^a_b ( {x}^{2} + 2x)dx}[/tex]

Before we get to Integral, you have to know Differentiation first. If you know how to differentiate a polynomial function then we are good to go in Integral!

We call the function that we are going to integrate as Integrand. Integrand is a function that's differentiated. In Integral, Integrating requires you to turn the function from differentiated to an original function.

For Ex. If the Integrand is x² then the original function is (1/3)x³ because when we differentiate (1/3)x³, we get x²

[tex] \large{f(x) = \frac{1}{3} {x}^{3} \longrightarrow f'(x) = {x}^{2} } \\ \large{f'(x) = 3( \frac{1}{3} ) {x}^{3 - 1} } \\ \large{f'(x) = {x}^{2} }[/tex]

So when we Integrate, make sure to convert Integrand as in original function. From the question, our Integrand is x²+2x. The function is in differentiated form. We know that x² is from (1/3)x³ and 2x comes from x²

[tex] \large{ f(x) = {x}^{2} \longrightarrow f'(x) = 2x} \\ \large{f'(x) = 2 {x}^{2 - 1} } \\ \large{f'(x) = 2x}[/tex]

Thus,

[tex] \large{ \int \limits^a_b ( {x}^{2} + 2x)dx} \\ \large{\int \limits^a_b ( \frac{1}{3} {x}^{3} + {x}^{2}) }[/tex]

Normally, if it's an indefinite Integral then we'd just put + C after (1/3)x³+x² but since we have a and b, it's a definite Integral.

[tex] \large{ \int \limits^b_a f(x)dx = F(b) - F(a)}[/tex]

Define F(x) as our anti-diff

From our problem, substitute x = a in then subtract with the one that substitute x = b

[tex] \large{ (\frac{1}{3}{a}^{3} + {a}^{2} ) - ( \frac{1}{3} {b}^{3} + {b}^{2}) }[/tex]

Simplify as we get:

[tex] \large \boxed{ \frac{1}{3}{a}^{3} + {a}^{2} - \frac{1}{3} {b}^{3} - {b}^{2}}[/tex]

Respuesta:

03:40 pm;

15: 40;

20 minutos antes de las 4 pm;

40 minutos después de las 3 pm

Explicación paso a paso:

20 minutos antes de las 4 de la tarde se pueden expresar de las siguientes formas:

Basado en un tiempo de 12 horas, podría escribirse tiene: 3: 40 pm

Usando el formato de reloj de 24 horas; donde 3 pm es equivalente a 15; por lo tanto, podría escribirse como: 15:40

Además, el tiempo se puede informar en palabras de la siguiente manera:

20 minutos antes de las 4 pm

También ;

40 minutos después de las 3 pm

Answer:

c

Step-by-step explanation:

because there is a 3 in front of the root it is a cubed root since

2= square- the two is invisible but its there

3= cubed

Answer:

15 days

Step-by-step explanation:

Don't you mean 1/3?  If so,

 5 cans

-------------------- = 15 days

1/3 can/daily

Answer: x=15

Step-by-step explanation: Since 3x+5 is equal to 50, you can write 3x+5=50, and 3x=45, x=15.

Answer:

20000/10

Step-by-step explanation:

Answer:

Step-by-step explanation:

2000%=2000/100=20/1