Can somebody classify this figure

The weight, y, in pounds, of kittens was tracked for the first 8 weeks after birth where t represents the number of weeks after

marshall27 [118]

Answer:

Step-by-step explanation:

The mentioned relationship for the weight, in pounds, of the kitten with respect to time, in weeks, is

$\hat y =1.7 +1.48t$

Weight of the kitten after 10 weeks

$\hat y =1.7 +1.48\times 10$

$\hat y =16.5$ pounds

This modeled equation is based on the observation of the early age of a kitten where the kitten is in its growth period, but in the early stage the growth rate in the weight of the kitten was the same but the growth of any living beings continues till the adult stage. So, after some time, in real life situation, this weekly change in weight will become zero, So, this model is not suitable to measure the weight of the kitten over the larger time period.

Here, t= 10 weeks is nearby the observed time period, so the linearly modeled equation can be used to predict the weight.

Hence, the weight of the kitten after 10 weeks is 16.5 pounds.

8 0

2 years ago

What is the answer please

Marina86 [1]

Hey there!

Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.

I hope this helps!

8 0

2 years ago

Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of

Mamont248 [21]

we have

$7=-2x^{2} +10x$

Factor the leading coefficient

$7=-2(x^{2} -5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$7-12.50=-2(x^{2} -5x+2.5^{2})$

$-5.50=-2(x^{2} -5x+2.5^{2})$

Divide both sides by $-2$

$2.75=(x^{2} -5x+2.5^{2})$

Rewrite as perfect squares

$2.75=(x-2.5)^{2}$

Taking the square roots of both sides (square root property of equality)

$x-2.5=(+/-)\sqrt{2.75}$

Remember that

$\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}$

$x-2.5=(+/-)\frac{\sqrt{11}}{2}$

$x=2.5(+/-)\frac{\sqrt{11}}{2}$

$x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}$

$x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}$

the answer is

The solutions are

$x=\frac{5+\sqrt{11}}{2}$

$x=\frac{5-\sqrt{11}}{2}$

5 0

1 year ago

Read 2 more answers

PLEASE HELP!

melamori03 [73]

Answer:

The answer is C.

Step-by-step explanation:

The $13 he started off with is the y-intercept or the initial value, which is the point where x=0.

The x-intercept, or rate of change, is $8.50.

The slope intercept formula is y=mx+b, where m=8.5 and b=13.

The equation for this problem is y=8.5x+13.

So the answer is C

4 0

2 years ago

Which number has the greatest absolute value? a. 0 b. 10 c. –10 d. –20?

Triss [41]

Keep in mind, absolute values are always positive, even when they say negative.That is because the absolute value of a number is just the distance that number is from 0. Example : -5 and 5...they are both 5 spaces from 0...doesn't matter if it is negative or positive, it is still 5 spaces.

|0| = 0
|10| = 10
|-10| = 10
|-20| = 20

so ur answer is : | -20 |

6 0

2 years ago