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3 years ago

For each of the transactions in items 1 through 5, indicate the effects on the accounting equation elements.

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

6 0

Answer:

Decrease I think.

Sorry if it is wrong.

Hope it helped :)

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South Celestial Pole is at the South Point

9966 [12]

Answer:

<em>The south celestial pole is the point in the sky directly above Earth's southern axis. It's the point around which the entire southern sky appears to turn. The height of the south celestial pole in your sky depends on your latitude.</em>

Step-by-step explanation:

3 0

3 years ago

During a cold front, the temperature in Denver, Colorado, dropped from 7 degrees C to -5 degrees C. What was the change in the t

Radda [10]

Answer:

-12 c

Step-by-step explanation:

4 0

3 years ago

I don’t understand this someone pls help

omeli [17]

To solve this, you need to plug in the numbers for <em>h</em>.

-4(-12) ≥ 8 48 ≥ 8 yes

-4(-7) ≥ 8 28 ≥ 8 yes

-4(-5) ≥ 8 20 ≥ 8 yes

-4(-3) ≥ 8 12 ≥ 8 yes

-4(-2) ≥ 8 8 ≥ 8 yes

-4(-1) ≥ 8 4 ≥ 8 no

-4(1) ≥ 8 -4 ≥ 8 no

-4(3) ≥ 8 -12 ≥ 8 no

-4(8) ≥ 8 -32 ≥ 8 no

Hope this helped!

7 0

3 years ago

The function below is written in vertex form or intercept form. Rewrite them in standard form and show your work.

lora16 [44]

Answer:

The standard form as $y=5x^2+30x+41$

Step-by-step explanation:

Given: A function which is written in vertex form or intercept form.

We have to re-write it in standard form that in terms of

Given $y = 5(x+3)^2-4$

Squaring using $(a+b)^2=a^2+b^2+2ab$ , we get,

$y=5(x^2+9+6x)-4$

Multiply 5 inside , we get,

$y=5x^2+45+30x-4$

Solving further , we get,

$y=5x^2+30x+41$

Thus , we have obtained the standard form as $y=5x^2+30x+41$

8 0

3 years ago

Read 2 more answers

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago