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3 years ago

Last 3 questions need to be answered!!

Mathematics

1 answer:

snow_tiger [21]3 years ago

4 0

2. just do inverse transformation to get pre-image. simply do T(4,-3) to given vertices.
you'll get A(-1,1)B(7,1)C(7,-2)D(-1,-2)

5. dont know.

10.option 1 and 4

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A square cement parking lot is being converted to a nature park. Each side of the parking lot is 100 meters long. The plan is to

irina [24]

Answer:

$A(t)=(100-8t)^{2}$

Step-by-step explanation:

Let

t-----> the number of years

A-----> the area of the cement parking lot

we know that

The function that represent this situation is

$A(t)=(100-8t)^{2}$

3 0

3 years ago

Help me please i dont get it.

Andre45 [30]

Answer:

$x > - \frac{1}{3}$

Step-by-step explanation:

$9(2 - x) - 4(2 - 3x) > 9$

$18 - 9x - 8 + 12x > 9$

$10 + 3x > 9$

$3x > 9 - 10$

$3x > - 1$

$\boxed{\green{x > - \frac{1}{3}}}$

3 0

3 years ago

Read 2 more answers

What is the sum of the geometric series

lbvjy [14]

If $S$ denotes the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ , then

$S=a+ar+ar^2+\cdots+ar^{n-3}+ar^{n-2}+ar^{n-1}$
$rS=ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1}+ar^n$
$\implies S-rS=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n$
$\implies (1-r)S=a(1-r^n)$
$\implies S=a=\dfrac{1-r^n}{1-r}$

Using summation notation, you have

$S=\displaystyle\sum_{x=1}^nar^{x-1}=\sum_{x=0}^{n-1}ar^x=a\dfrac{1-r^n}{1-r}$

In this case, you have $a=2$ , $r=\dfrac12$ , and $n=16$ . So the value of the sum is

$\displaystyle\sum_{x=0}^{15}2\left(\frac14\right)^x=2\dfrac{1-\left(\frac14\right)^{16}}{1-\frac14}\approx2.67$

Rounded to the nearest whole number, the answer would be 3.

4 0

4 years ago

Read 2 more answers

Can someone help me answer this question

ch4aika [34]

What? Need answers. Plus, I can tell that your are doing this in class. Bad idea.

3 0

3 years ago

What do tge coefficents on the second and first degree terms in the original equation tell you about the arrows flight

kati45 [8]

-4:9 - This is half of the local gravitational constant, -9.8 m/s^2.

29:4 - The initial velocity of the arrow as it was shot upward was 29.4 m/s <u>(approximately 66 mph)</u>

6 0

3 years ago