Someone pls help me with this plsss

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Ierofanga [76]

The area of the triangle is $2\sqrt{3}$ = 3.4641 square feet.

Step-by-step explanation:

Step 1:

According to the Pythagorean theorem, the square of the hypotenuse will be equal to the sum of the squares of the other two sides.

In the given triangle, the hypotenuse is the side that is represented by c. The other sides on the triangle are represented by the values a and b.

Step 2:

Assume the other side of the triangle measures x units.

So according to the Pythagorean theorem,

$4^{2} = (2\sqrt{3} )^{2} + x^{2} , x^{2} = 4^{2} - (2\sqrt{3} )^{2} .$

$x^{2} = 16 - 12 =4, x =2.$

So the other side of the triangle, b measures 2 feet.

Step 3:

The area of a triangle is half the product of its base length and height.

The base length of this triangle is $2\sqrt{3}$ feet and the height is 2 feet.

The area of the triangle $= \frac{1}{2} (2\sqrt{3} )(2) = 2\sqrt{3} = 3.4641.$

The area of the triangle is $2\sqrt{3}$ = 3.4641 square feet.

3 0

2 years ago

Find the sum of the geometric series for which a = 160, r = 0.5, and n = 6.

harina [27]

$\\ \sf\longmapsto S_n=\dfrac{a(1-r^n)}{1-r}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.5^6)}{1-0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(1-0.015625)}{0.5}$

$\\ \sf\longmapsto S_n=\dfrac{160(0.984378)}{0.5}$

$\\ \sf\longmapsto S_n=80(0.984378)$

$\\ \sf\longmapsto S_n=78.75$

6 0

2 years ago

Read 2 more answers

Please help I’m struggling rlly bad w this

Rzqust [24]

if we convert 6/8 to a smaller fraction it became 3/4. then, if we convert 27/36, it became 3/4. so, 6/8 os equivalent with 27/36. so the answer is 216=216

7 0

2 years ago

Refer to the following information from a frequency distribution for "heights of college women" recorded to the nearest inch:The

valkas [14]

Answer: Therefore, Option 'A' is correct.

Step-by-step explanation:

Since we have given that

Two class midpoints are 62.5 and 65.5

So, the class interval would be

$65.5-62.5=3$

So, the limits of lower class are $x_1\ and\ x_2$

The limits of upper class are $x_2\ and\ x_3$

So, it becomes,

$\dfrac{x_1+x_2}{2}=62.5\\\\x_1+x_2=62.5\times 2=125$

and

$\dfrac{x_2+x_3}{2}=65.5\\\\x_2+x_3=65.5\times 2=131$

Since we have given that interval is 3.

So, $x_2-x_1=3\ and\ x_3-x_2=3$

So, by solving all the questions, we get that

$x_2+x_1=125\\\\x_2-x_1=3\\\\-------------------------\\\\2x_2=128\\\\x_2=64\\\\x_2-x_1=3\\\\64-x_1=3\\\\x_1=61\\\\and\\\\x_3-x_2=3\\\\x_3-64=3\\\\x_3=67$

Hence, the class limits would be

(61,64) and (64,67).

And the class limits of lowest class are 61 and 64.

Therefore, Option 'A' is correct.

4 0

3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Travka [436]

Answer:

a) 4x - 8y

b) 4 × x - 8 × y

c) 4x - 8y is the same as 4 × x - 8 × y because if there is no simbol then it is always multiplication

Step-by-step explanation:

a) start with the parentheses

1/3 (6x - 12y)

multiply 6x and -12y by (1/3)

2x - 4y

so now the equation is 5x -3x -4y + 2x -4y

add together the like terms

5x -3x + 2x = 4x

-4y -4y = -8y

so we have 4x - 8y

b) im not good with factored form but i believe this is right

take 4x and separate them by turning it into 4 × x and same for -8y turn it into -8 × y

c) 4x - 8y is the same as 4 × x - 8 × y because if there is no simbol then it is always multiplication

im not sure how to prove them equal to the original expression i dont really remember that but i hope this helped

3 0

2 years ago