A plant starts with 1 branch. Every year,
each branch becomes 3 branches. A
sketch of the plant for the first 3 years is
shown. How many branches will the plant
have in year 10?
TYP
I V
Year 1
Year 2
Year 3
How many branches would the plant
have in year 10 if the plant ad
5 branches the first year? (Each branch
still becomes 3 branches every year.

Answer:

The last one

Step-by-step explanation:

Look at A and A'

It moved 4 units to the right and 1 unit down

The same with B, B' and C, C'

The perimeter of the isosceles right triangle comes to be 6+3√2.

Given that the length of one leg of an isosceles right triangle = 3 ft

A right-angle triangle having sides other than hypotenuse equal to each other is called an isosceles right triangle.

So, The length of the other leg of an isosceles right triangle = 3 ft

From Pythagoras theorem

Hypotenuse²=Perpendicular²+Base²

So, in the given isosceles right triangle

Hypotenuse²=Leg 1²+Leg2²

Leg 1=Leg2 = 3 feet

Hypotenuse² =  3²+3²

Hypotenuse = 3√2 ft.

So, the perimeter of the triangle = 3+3+ 3√2

The perimeter of the triangle =6+3√2

Therefore, The perimeter of the isosceles right triangle comes to be 6+3√2.

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Answer:

6 + 3 StartRoot 2 EndRoot ft

Step-by-step explanation:

Edge2022

Answer:

2 that's what soup calculated says

Answer:

(2,11 2/3)  or (2,(35/3))

Step-by-step explanation:

5x-3y=-25

For x = 2:  

5(2)-3y=-25

10-3y=-25

-3y=-35

y = (35/3) or 11 2/3

(2,11 2/3)

Answer:

X = -2

Y= 5

You use:

5x-3(5)=-25

and

5(-2)-3y=-25

The central angle (127 degrees) is the angle at point K

The measures of JL and JML are 127 and 233 degrees, respectively

From the complete question, we have:

JL = 127 degrees.

The sum of angles at a point is 360 degrees

So, we have:

JML + 127 = 360

Subtract 127 from both sides

JML = 233

Hence, the measures of JL and JML are 127 and 233 degrees, respectively

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The required values are :

[tex]\qquad \sf  \dashrightarrow \:x = 4[/tex]

and

[tex]\qquad \sf  \dashrightarrow \:y = 1[/tex]

Check the attachment for solution ~

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Answer:

A

Step-by-step explanation:

The zeros are the roots or x-intercepts

They are located at coordinate points (0, 0) and (6, 0)

This means the answer is A

Hope it makes sense!

Answer:

A. x = 0 and x = 6

Step-by-step explanation:

So, the zeroes of a quadratic are the parts of it that touch the x-axis. When you find those parts, take their x value and you will get the zero.

So, in this parabola, it touches the x-axis at (0, 0) and (6, 0).

The x-values of these are 0 and 6, respectively.

Function: y = 5 sin (2(x))+ 12

Find y-intercept:

y = 5 sin 2(0)+ 12

y = 12

​[tex]\sf Y\:Intercepts}:\:\left(0,\:12\right)[/tex]

→ Formula for maximum: M = A + |B|

Maximum:

12 + |5|

17

When y = 17

[tex]\rightarrow \sf 17 = 5 sin (2(x))+ 12[/tex]

[tex]\rightarrow \sf \sf 5 sin (2(x)) = 5[/tex]

[tex]\rightarrow \sf sin (2(x)) = 1[/tex]

[tex]\rightarrow \sf 2x = sin^{-1}(1)[/tex]

[tex]\rightarrow \sf 2x = 90^{\circ \:}, \ \ 450^{\circ \:}[/tex]

[tex]\rightarrow \sf x = 45^{\circ \:}, \ \ 225^{\circ \:}[/tex]

maximum: ( 45° , 17 ), (225° , 17), .....

==========================================================

→  Formula for minimum: m = A ‐ |B|

Minimum:

12 - |5|

7

When y = 7

[tex]\rightarrow \sf 7 = 5 sin (2(x))+ 12[/tex]

[tex]\rightarrow \sf 5 sin (2(x)) = -5[/tex]

[tex]\rightarrow \sf 2(x) = sin^{-1}(-1)[/tex]

[tex]\rightarrow \sf x = -45^{\circ \:} , \ \ 135^{\circ \:}[/tex]

minimum: ( -45°,7), (135°, 7), .....

Repeat the same process for finding more values on the x-axis, or just follow the trend of the curve from the points found and sketch the graph easily.

[tex]\sf Domain\:\left(-\infty \: < x < \infty )[/tex]

[tex]\sf Range : 7\le \:f\left(x\right)\le \:17[/tex]

Sketched below:

Answer:

0

Step-by-step explanation:

it less than 4 so u can't pick 4,so it is any number less than 4

Answer:

B) 0

Step-by-step explanation:

The open circle on 4 and the arrow pointing to the left means x < 4

x < 4 means that x is smaller than 4

4 = 4   incorrect as this says 4 is equal to 4

0 < 4   CORRECT as zero is smaller than 4

5 > 4   incorrect as 5 is bigger than 4

4.5 > 4   incorrect as 4.5 is bigger than 4

Answer:

amplitude = 2

Step-by-step explanation:

the amplitude is the distance from the x- axis(midline ) to the maximum value , that is

amplitude = 2

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Option B ( 2 ) ✓

OR,

[tex] \sf{amplitude = distance \: from \: undisturbed \: position \div centre \: line \: to \: through \: or \: peak \: }[/tex]

[tex]\purple{\rule{15mm}{2.9pt}} \red{\rule18mm{2.5pt}} \orange{ \rule18mm{2.5pt}}[/tex]

[tex]\sf{\:мѕнαcкεя\: ♪...}[/tex]

Answer:

please rate and mark as brainlest.

Step-by-step explanation:

(i) 1/20

(ii)

a.708.22

b.8635.4

c.147.392

please rate and mark as brainlest.

Answer:

3.75?

Step-by-step explanation:

I dont know if this is what youre looking for but I guess?

Answer:

(2, -1)

Step-by-step explanation:

x = 2, y = −1

Answer:

Step-by-step explanation:

Put the values of x to the equation y = 2x + 2 and calculate the value of y:

x = 2

y = 2 · 2 + 2 = 4 + 2 = 6

x = 5

y = 2 · 5 + 2 = 10 + 2 = 12

x = 6

y = 2 · 6 + 2 = 12 + 2 = 14

x = 8

y = 2 · 8 + 2 = 16 + 2 = 18

Answer:

Given inequality:  [tex]x+8y\leq 24[/tex]

Part A

Solid line: ≤ or ≥

Dashed line: < or >

Therefore, the line will be solid.

Part B

Shading above the line: y > or y ≥

Shading below the line: y < or y ≤

Therefore, shade below the line.

Part C

Make y the subject to determine where to shade.

Given inequality:  

[tex]x+8y\leq 24[/tex]

Subtract x from both sides:

[tex]8y\leq 24-x[/tex]

Divide both sides by 8:

[tex]y\leq 3-\dfrac18x[/tex]

[tex]\\ \rm\Rrightarrow x+8y\leqslant 24[/tex]

#Part A

#Part B

Put (0,0)

Satisfied

Shading below the line and towards origin

#Part C

turn into y=mx+b form

Answer:

Add 5 to both sides

Step-by-step explanation:

x-5=16

+5  +5

x=21

Answer:

Addition

Step-by-step explanation:

x - 5 = 16

x - 5 + 5 = 16 + 5

x = 21

Answer: 0.0139358  

Step-by-step explanation: 999,999,999,999,999,999,999,999,999,999,999% sure!

Answer:Examples of quadratic equations are: 6x² + 11x – 35 = 0, 2x² – 4x – 2 = 0, 2x² – 64 = 0, x² – 16 = 0, x² – 7x = 0, 2x² + 8x = 0 etc. From these examples, you can note that, some quadratic equations lack the term “c” and “bx.”

Step-by-step explanation:

A quadratic equation in the variable x is an equation of the form ax^2 + b x + c = 0, where a, b, c are real numbers, a ≠ 0.

Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations.

The standard form of a quadratic equation is;

Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax^2 + bx + c where a, b, c, ∈ R and a ≠ 0.

These are the following examples of the quadratic equation;

x^2-2x

x^2+5x+6

x^2-5x

Any equation of the form p (x) = 0, where p (x) is a polynomial of degree 2, is a quadratic equation.

If we write the terms of p (x) in decreasing order of their degrees, then we get the standard form of the equation.

Learn more about quadratic equations here;

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Answer:

A. 2086 yd²

Step-by-step explanation:

SA = 2lw + 2wh + 2lh
SA = 2*17*20 + 2*20*19 + 2*17*19
SA = 680 + 760 + 646
SA = 2086
Hope this helps :)

The total harvested area is 232 acres .

To determine the number of acres harvested, we simply have to divide the bushels of soybean by the crop yield.

Mathematically we can write it as,

The total land area = bushels of soybean / bushels per acre

Substituting the known values,

[tex]Total land area = 9396 / 40.5 = 232[/tex]

Therefore, 232 acres were harvested.

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Answer:The total harvested area is 232 acres

Step-by-step explanation:

Answer:

can't explain without you arranging your mathematical expression

Answer:

8 (or 8.4)

Step-by-step explanation:

First, the problem states the words: Distinct and positive.

That means that the smallest number we can use is 0.

We can make 10 blanks for 10 numbers for us to fill in.

_ _ _ _ _ _ _ _ _ _

Since 10 is an even number, and the median is not one single number, it will be the middle of the two numbers.

In this case, our two numbers are the 5th and 6th blank.

Any two numbers can be used, as long as they are the same actual value from 10.

Let us first put in the numbers we can, which are the numbers before the two middle blanks.

0,1,2,3, _ _ _ _ _ _

To find the median between two numbers, we can do:

(a+b) divided by 2 = median

We can use the smallest following number, 4. Then the 6th number will have to be 16 for 10 to be in the middle.

Also try the largest possible number for the 5th blank, 9. Then the 6th number will be 11.

4+16 and 9+11 both equal 20, and 20 divided by 2 is 10. So both of these work.

Now let's place the other numbers in for these two equations.

0,1,2,3,4,16,17,18,19,20

0,1,2,3,9,11,12,13,14,15

If we add the numbers of each together, we get:

0,1,2,3,4,16,17,18,19,20=100

0,1,2,3,9,11,12,13,14,15= 80

If we now divide each sum by 10 (to find the average) we get:

10

8

Since both of these were the most we could go, one with the 5th number as small as possible and one with the 5th number as large as possible.

Since the smaller answer we got was 8, the answer must be 8.

(I believe 0 is a positive integer because it doesn't carry a negative sign. If 0 is not a positive integer, the answer is 8.4. Use the same process

Answers:

5.) [tex]\displaystyle \pm{1\frac{2}{3}}[/tex]

4.) [tex]\displaystyle 2[/tex]

3.) [tex]\displaystyle \pm{1\frac{2}{3}}[/tex]

2.) [tex]\displaystyle 2[/tex]

1.) [tex]\displaystyle \pm{2}[/tex]

Step-by-step explanations:

5.) [tex]\displaystyle \frac{\sqrt{75}}{\sqrt{27}} \hookrightarrow \sqrt{2\frac{7}{9}} \hookrightarrow \frac{\pm{5\sqrt{3}}}{\pm{3\sqrt{3}}} \\ \\ \boxed{\pm{1\frac{2}{3}}}[/tex]

4.) [tex]\displaystyle \frac{\sqrt[5]{96}}{\sqrt[5]{3}} \hookrightarrow \sqrt[5]{32} \hookrightarrow \frac{2\sqrt[5]{3}}{\sqrt[5]{3}} \\ \\ \boxed{2}[/tex]

3.) [tex]\displaystyle \frac{\sqrt{125}}{\sqrt{45}} \hookrightarrow \sqrt{2\frac{7}{9}} \hookrightarrow \frac{\pm{5\sqrt{5}}}{\pm{3\sqrt{5}}} \\ \\ \boxed{\pm{1\frac{2}{3}}}[/tex]

2.) [tex]\displaystyle \frac{\sqrt[3]{24}}{\sqrt[3]{3}} \hookrightarrow \sqrt[3]{8} \hookrightarrow \frac{2\sqrt[3]{3}}{\sqrt[3]{3}} \\ \\ \boxed{2}[/tex]

1.) [tex]\displaystyle \frac{\sqrt{8}}{\sqrt{2}} \hookrightarrow \sqrt{4} \hookrightarrow \frac{\pm{2\sqrt{2}}}{\pm{\sqrt{2}}} \\ \\ \boxed{\pm{2}}[/tex]

I am joyous to assist you at any time.

Step-by-step explanation:

Large Jar 540g for 2.16

540/2.16 = 250

(Always divide by the smallest ratio)

= 250g per 1£

Small Jar 360g for 1.37

360/1.37=262.77g

= 262.77g per 1£

Small Jar is the best value for money, more grams, same price.

Answer:

10:48 pm will be your answer

Answer:

[tex]\sf \dfrac{x}{8}-22[/tex]

Explanation:

[tex]\rightarrow \sf \left(\dfrac{7}{8}x-4\right)-3\left(\dfrac{1}{4}x+6\right)[/tex]

remove parenthesis

[tex]\sf \rightarrow \dfrac{7}{8}x-4-\dfrac{3x}{4}-18[/tex]

expand

[tex]\rightarrow \sf \dfrac{7x}{8}-\dfrac{3x}{4}-22[/tex]

simplify

[tex]\rightarrow \sf \dfrac{x}{8}-22[/tex]

Answer:

[tex]\frac18x-22[/tex]

Step-by-step explanation:

Given expression:

[tex](\frac78x-4)-3(\frac14x+6)[/tex]

Expand:

[tex]\implies \frac78x-4-\frac34x-18[/tex]

Gather like terms:

[tex]\implies \frac78x-\frac34x-4-18[/tex]

Change 3/4 into eighths:

[tex]\implies \frac78x-\frac68x-4-18[/tex]

Combine like terms:

[tex]\implies \frac18x-22[/tex]

Answer:

( - y - 3 , - x - 1 )

Step-by-step explanation:

( x , y ) ----> ( - y , x ) ----> ( - y - 3 , x + 1 ) ----> ( - y - 3 , - x - 1 )