Write a recursive rule for the sequence.<br><br> 7, 6, 13, 19, 32, ...

Describe the end behavior of the graphed function

Troyanec [42]

You have to consider the “ends” of the x-axis, the far right (for infinitely large values of x) and left (for infinitely small values of x) of the graph.

From the diagram above you can see that:

When $x\rightarrow -\infty,$ then $y=f(x)\rightarrow 1$ (notice that as the values of x get smaller and smaller, the graph gets closer and closer to the line y=1);
When $x\rightarrow \infty,$ then $y=f(x)\rightarrow 1.$ (notice that as the values of x get larger and larger, the graph gets closer and closer to the line y=1).

Answer: correct choice is D.

5 0

3 years ago

Which point is not on the graph of y =2x + 3?

Nostrana [21]

Answer:

i'm pretty sure its D

Step-by-step explanation:

if you make a graph yourself you'll see all the rel;relationship match except for d

3 0

2 years ago

Translate to a proportion and solve.<br> 148.5 is 59 1/5% of what?

lesantik [10]

The actual value is 250.84 .

Formula

Percentage calculation =(given value/total value)*100

let the actual value is "x"

percentage we known is 59 1/5% (59.2%) and 59.2% of x is 148.5.

for the percentage,

x*59.2% =148.5

x*59.2/100=148.5

x=(148.5/59.2)*100

x=(2.5084)*100

x=250.84

The actual value is 250.84 .

Learn more about the percentage here:

brainly.com/question/14091961

#SPJ1

5 0

1 year ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

Write the conversion factor for converting meters to centimeters

posledela

Since we are converting meters to centimeters we are converting large to small.

Expect the answer to be a larger number. Here are some example of large to small:
2 weeks (larger unit) = 14 days (smaller unit) 14 is larger than 2
4 days (larger unit ) = 96 hours (smaller unit) 96 is larger than 4
Since there are 100cm in a meter, when we change meters to cm, we need a bigger number. The conversion factor is 100 and should be used as a multiplier.
Example: 47m x 100 = 4700 cm

7 0

3 years ago

Write a recursive rule for the sequence. 7, 6, 13, 19, 32, ...