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snow_tiger [21]

2 years ago

5

Translate the figure 2 units left and 4

1 answer:

egoroff_w [7]2 years ago

4 0

The answer is in the picture I’ll send u a link phs.dcsdk12.org the answer is in there good luck!

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How do i find the answer to this:<br><br> exponential equation 2^(x-4)=3^(2x)

qaws [65]

What area of math is this,I may be able to help?

4 0

2 years ago

Write in exponential notation: 301=

dalvyx [7]

Answer:

$3.01 * 10^2$

Step-by-step explanation:

=)

Hope this helps!

4 0

2 years ago

Solving Rational equations. LCD method. Show work. Image attached.

posledela

$(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12$

So in order to get all the fractions to have a common denominator, we need to multiply $\dfrac k{k-2}$ by $\dfrac{k-6}{k-6}$ , and $\dfrac1{k-6}$ by $\dfrac{k-2}{k-2}$ :

$\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}$

$\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}$

Now,

$\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}$

As long as $k\neq2$ and $k\neq6$ (which we can't have because otherwise $k^2-8k+12=0$ ), we can cancel $k^2-8k+12$ in the denominators on both sides:

$4=(k^2-6k)+(k-2)$

$4=k^2-5k-2$

$0=k^2-5k-6$

We can factorize the right side:

$0=(k-6)(k+1)$

which tells us that $k=6$ and $k=-1$ are solutions.

4 0

3 years ago

Bessy has 6 times as much money as Bob, but

poizon [28]

Answer:

Before: Bob, $4; Bessy, $24
After: Bob, $10; Bessy, $30

Step-by-step explanation:

Let x represent the amount of money Bob starts with. Then Bessy starts with 6x. After they have each earned $6, the relationship of their amounts is ...

(6x +6) = 3(x +6)

6x +6 = 3x +18

3x = 12

x = 4

Bob starts with $4; Bessy with $24.

After earning $6, Bob has $10; Bessy has $30.

8 0

3 years ago

Help me 2x-4y-3(7x-2y)

Zinaida [17]

Let's simplify the above expression :

$2x - 4y - 3(7x - 2y)$

$2x - 4y - 21x + 6y$

$- 19x + 2y$

4 0

2 years ago

Read 2 more answers