All categories

3 years ago

Find the area of the trapezoid

Mathematics

1 answer:

daser333 [38]3 years ago

4 0

Answer:

52√3 ft^2

Step-by-step explanation:

The area of a trapezoid is the average of the bases times the height

We have enough information to find these parts.

The average of the bases is 11+15 /2 =13

The height is 4√3

So the answer is 13 * 4√3 = 52√3

Please vote me for brainliest i really need it, i hope i helped you. :)

You might be interested in

Directions:Use the product rule to simplify the following monomials.

nadezda [96]

Answer: $-12m^3n^3$

Step-by-step explanation:

$\frac{1}{4}\left(8mn\right)\left(-6m^2n^2\right)$

$=-\frac{1}{4}\cdot \:8mn\cdot \:6m^2n^2$

$=-\frac{1}{4}\cdot \:48mnm^2n^2$

$=-\frac{1}{4}\cdot \:48m^3nn^2$

$-\frac{1}{4}\cdot \:48m^3n^3$

$=-12m^3n^3$

6 0

3 years ago

30 points to anyone that can answer correctly in probability.

IrinaVladis [17]

Formula: P(A or B)=P(A)+P(B)-P(A or B)

p(c)=7/18

p(s)=6/18

p(c or s)=2/18

0.39+0.33-0.11=

answer:0.61

4 0

3 years ago

Read 2 more answers

I really need help would please like help thanks

ASHA 777 [7]

Hey there!

A reciprocal might be an unfamiliar vocab word, but the concept is really quite simple.

It's just a flipped fraction! Multiplying the flipped fraction by it's reciprocal gets you one, and that's how you check your answer.

So we have:

Reciprocal of 9/10 is 10/9, because 9/10 times 10/9 = 90/90 = 1

Reciprocal of 1/11 is 11/1, because 1/11 times 11/1 = 11/11 = 1

Reciprocal of 10(remember, it's 10/1) is 1/10, because 10/1 times 1/10 = 10/10 = 1

Hope this helps!

4 0

2 years ago

The square of a number is 3 less than four times the number. Which values could be the number? 0 0 -3 O 1

grigory [225]

Answer:

3 and/or 1.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago