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2 years ago

10

HELPPPPPPPPPPPP!!!!!

2 answers:

kykrilka [37]2 years ago

6 0

9=3^2

3^(2x)=(3^2)^(3x-4)
3^(2x)=3^(6x-8)
2x=6x-8
0=4x-8
8=4x
2=x

Gennadij [26K]2 years ago

5 0

Answer:

9=3^2

3^(2x)=(3^2)^(3x-4)

3^(2x)=3^(6x-8)

2x=6x-8

0=4x-8

8=4x

2=x

Step-by-step explanation:

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If x = 3 units, y = 13 units, and h = 7 units, then what is the area of the trapezoid shown above?

Whitepunk [10]

(x)(y)(h)/ 2.
=136.5 unit

3 0

2 years ago

Solve this equation<br> 16(1/4x-1/2)>24-2x

jekas [21]

Answer:

x>14

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

−4>−2x+24

Step 2: Flip the equation.

−2x+24<−4

Step 3: Subtract 24 from both sides.

−2x+24−24<−4−24

−2x<−28

Step 4: Divide both sides by -2.

−2x−2<−28−2

x>14

5 0

2 years ago

Read 2 more answers

Sarah kicked a ball in the air. The function

Aleksandr-060686 [28]

Answer: The ball hits the ground at 5 s

Step-by-step explanation:

The question seems incomplete and there is not enough data. However, we can work with the following function to understand this problem:

$f=30 t- 6t^{2}$ (1)

Where $f$ models the height of the ball in meters and $t$ the time.

Now, let's find the time $t$ when the ball Sara kicked hits the ground (this is when $f=0 m$ ):

$0=30 t- 6t^{2}$ (2)

Rearranging the equation:

$6t^{2}-30 t=0$ (3)

Dividing both sides of the equation by $6$ :

$t^{2}-5 t=0$ (4)

This quadratic equation can be written in the form $at^{2}+bt+c=0$ , and can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=1$

$b=-5$

$c=0$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(0)}}{2(1)}$ (6)

Solving we have the following result:

$t=5 s$ This means the ball hit the ground 5 seconds after it was kicked by Sara.

6 0

2 years ago

Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

<u>Properties of Logarithms</u>

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

5 0

2 years ago

Simplify 2[(9 - 5)^5 / 8]

Alenkinab [10]

2[(9 - 5)^5/8] = 2[4^5/8] = 2[1024/8] = 2[128] = 256

5 0

3 years ago