Pleaseeeee helpppppppppppppppppppppppp

Answer:

36

Step-by-step explanation:

So to find the last angle first. The formula is a power 2+b power 2=c power 2

So 9 power 2 is 81 and 12 power 2 is 144

144+81=225

the square root of 225 is 15. So x is 15

No we just have to do 9+12+15=36.

36 is the perimeter

Answer:

Y =-25x +225

Step-by-step explanation:

First you can count the gradient first. Use then you can count y intercept

Answer:

Step-by-step explanation:

7 x 5 = 35

35/2 = 17.5

[tex]17.5ft^{2}[/tex]

The area of the given triangle is 17.5 ft²

Area is defined as the total space taken up by a flat (2-D) surface or shape of an object.

Given a triangle with base=7ft and height =5ft

We know that area of a triangle = 1/2 × base × height

Therefore, the area of a triangle = 1/2 (7)(5) = 17.5 ft²

Hence, the area of the given triangle is 17.5 ft²

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Answer:

25cm

Step-by-step explanation:

one the possibility 36 - 11 =25 the answer is 25

A right angled triangle with its side length is given. The sides of the triangle are 12ft , 13ft and 5ft. And we need to find out the value of cot A .

Here :-

We know :-

[tex]\tt\implies cot \theta = base / perpendicular [/tex]

Substitute the respective values :-

[tex]\tt\implies cot \theta = base / perpendicular\\\\\tt\implies\boxed{\red{\tt cot A = 12/5 }}[/tex]

Hence the required answer is 12/5.

Answer:

120

Step-by-step explanation:

reduce the fraction from

[tex]48 \times \frac{250}{100 } [/tex]

to

[tex]48 \times \frac{5}{2} [/tex]

then multiply and you get 24 × 5 which is 120

The interquartile range(IQR) of the recorded weight of ken is calculated as 17 pounds.

The interquartile range(IQR) = third quartile(Q3) - first quartile(Q1)

The third quartile of in the box plot given is : 56 pounds

The third quartile of the data as represented in the box plot given :

39 pounds

The interquartile range(IQR) of ken's data = 56 - 39

= 17 pounds

Hence, The interquartile range(IQR) of the recorded weight of ken is calculated as 17 pounds.

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Answer:

no thanks for your number

Step-by-step explanation:

Have a great day

The volume of a pyramid with a square base is 17.6 cubic centimeter.

Given the following data;

To find the volume of a pyramid with a square base;

Mathematically, the volume of a pyramid with a square base is given by the formula;

[tex]V = A(\frac{h}{3})[/tex]

Where;

First of all, we would determine the length of sides of the square base from its perimeter;

[tex]Perimeter \ of \ square = 4s\\\\9.9 = 4s\\\\s = \frac{9.9}{4}[/tex]

s = 2.475 cm

Next, we would determine the area of the square;

[tex]Area \ of \ a\ square = s^{2} \\\\Area \ of \ a\ square = 2.475^{2}[/tex]

Area of a square = 6.126 cm

Now, we would use the above value to find the volume of a pyramid with a square base.

Substituting the values into the formula, we have;

[tex]V = 6.126*(\frac{8.6}{3})\\\\V = 6.126*2.867[/tex]

Volume, V = 17.6 cubic centimeter

Therefore, the volume of a pyramid with a square base is 17.6 cubic centimeter

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Answer:

π2,√80,12

Step-by-step explanation:

π2= 6.28

√80= about 8

12=12

Hope I could help!

Answer:

D) y = -4/3x + 5

Step-by-step explanation:

Can be checked with a calculator or by replacing the variables into the equation.

the total balloon are used is 2736

Answer:

40 because 40 percent of 100 is 40. You can just multiply the number by the percentage $100x40%

Answer:

I'm sorry, I can help you with only this one :( ∠F = 44°

Step-by-step explanation:

∠F = 180° - (46° + 90°) = 44°

Answer:

x = [tex]\frac{1}{2}[/tex] , x = 1

Step-by-step explanation:

Given

[tex]\frac{2x}{x+3}[/tex] + [tex]\frac{3}{x-3}[/tex] = [tex]\frac{8}{x^2-9}[/tex] ← denominator is a difference of squares

[tex]\frac{2x}{x+3}[/tex] + [tex]\frac{3}{x-3}[/tex] = [tex]\frac{8}{(x+3)(x-3)}[/tex]

Multiply through by (x + 3)(x - 3) to clear the fractions

2x(x - 3) + 3(x + 3) = 8 ← distribute left side and simplify

2x² - 6x + 3x + 9 = 8

2x² - 3x + 9 = 8 ( subtract 8 from both sides )

2x² - 3x + 1 = 0 ← in standard form

(2x - 1)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = [tex]\frac{1}{2}[/tex]

x - 1 = 0 ⇒ x = 1

[tex]\huge\text{Hey there!}[/tex]

[tex]\large\textsf{2h - 3 }\mathsf{>}\large\textsf{ 15}[/tex]

[tex]\large\text{ADD 3 to BOTH SIDES}[/tex]

[tex]\large\textsf{2h - 3 + 3 }>\large\textsf{ 15 + 3}[/tex]

[tex]\large\text{CANCEL out: \textsf{-3 + 3} because that gives you 0}[/tex]

[tex]\large\text{KEEP: \textsf{15 + 3} because it helps you solve for your h-value}[/tex]

[tex]\large\textsf{15 + 3 = \boxed{\large\textsf{\bf 18}}}[/tex]

[tex]\large\text{NEW EQUATION: \textsf{2h} }\mathsf{>}\large\textsf{ 18}[/tex]

[tex]\large\text{DIVIDE 2 to BOTH SIDES}[/tex]

[tex]\mathsf{\dfrac{2h}{2}>\dfrac{18}{2}}[/tex]

[tex]\large\text{CANCEL out: }\mathsf{\dfrac{2}{2}}\large\text{ because it gives you 1}[/tex]

[tex]\large\text{KEEP: }\mathsf{\dfrac{18}{2}}\large\text{ because it gives you your h-value is being compared to}[/tex]

[tex]\large\text{NEW EQUATION: }\mathsf{h > \dfrac{18}{2}}[/tex]

[tex]\mathsf{\dfrac{18}{2}}[/tex]

[tex]\mathsf{= 18\div 2}[/tex]

[tex]=\boxed{\mathsf{\large\textsf {\bf 9}}}[/tex]

[tex]\boxed{\boxed{\large\textsf{Answer: }\mathsf{\bf h > 9}}}\huge\checkmark\\\\\boxed{\boxed{\rm \bf It\ it\ an\ O P E N E D\ circle\ shaded}}\\\boxed{\boxed{\rm \bf to\ the\ right\ starting\ at\ point\ 9}}\checkmark[/tex]

[tex]\large\text{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 > 15}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 > 15}[/tex]

[tex] \quad \quad \quad \quad\tt{2h - 3 + 3 > 15 + 3}[/tex]

[tex]\quad \quad \quad \quad\tt{2h \: \: \cancel{ \color{red}- 3 + 3} \: > 15 + 3}[/tex]

[tex]\quad \quad \quad \quad\tt{2h > 18}[/tex]

[tex]\quad \quad \quad \quad\tt{ \frac{2h}{2} > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ \frac{ \cancel{ \color{red}2}h}{ \cancel{ \color{red}2}} > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ h > \frac{18}{2} }[/tex]

[tex]\quad \quad \quad \quad\tt{ h > 9}[/tex]

[tex]\quad \quad \quad \quad \boxed{\tt { \color{green}h > 9}}[/tex]

________

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Answer:

H

Step-by-step explanation:

For F  - Median For Class 1 is 52.5 Median for Class 2 is 75 ( False )

For G - Range for class 2 is 75 Range for class 1 is 45.9 ( False )

For H - IQR for class 1 is 30.45 IQR for class 2 is 60 ( True )

For J - Minimum for Class 2 is 30 Minimum for Class 1 is 30 ( False )

Answer:

1. -1/2 1/2 1/4

2. -3.5 -3.6 -3.7

3. -1/10 -1/11 -1/12

hope this helps

have a good day :)

Step-by-step explanation:

Answer:

landslide

Step-by-step explanation:

Answer:

I believe that the answer might be the second one

Volcanic Eruption

Answer: look at the picture

Step-by-step explanation: Hope this help :D

Answer:

I think ir is 2 -2

Step-by-step explanation:

hope it helps

Step-by-step explanation:

put 6x-10 in y=x-4

so

y = (6x-10) - 4

y = 6x-14

Answer:

the second store has the better deal.

Step-by-step explanation:

12 ÷ 2 = 6

7.5 × 6 = 45

20.2 < 45