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3 years ago

What is the area of this figure?

Mathematics

2 answers:

zloy xaker [14]3 years ago

8 0

First calculate the area of the parallelogram, which is 20 x 16, and you get 320, then add the triangle at the top, which is (20 x 8) divided by 2. where you get 80. so add both of them together, and the answer should be 400.

bagirrra123 [75]3 years ago

3 0

The area would be 400 inches:

To find the area this figure, find the areas in smaller portions.
The triangle on the top would be base times height, divided by 2
20 * 8 = 80

Then the parallelogram's area would be base times height:
20 * 16 = 320

Add both of the numbers together and you get 400 inches!

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Using Substitution solve: y=x+5<br> 4x+y=20

Vladimir [108]

Answer:

Step-by-step explanation:

So first, plugin y to your equation, it should look like this: 4x+x+5+20. Now, we can solve this. Next, we have two x's so we can combine them. The equation will now look like this: 5x+5=20. Now, we have an extra 5 laying around that we need to get rid of, so we can subtract that 5 to both sides of the equation and we get this: 5x-5=20-5 which ends up being 5x=15. Now, we just divide both ides by 5 and since 15 divided by 5 is 3, we get: x=3

4 0

3 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago

Given that D is the midpoint of AB and B is the midpoint of AC which statement must be true a) AD= 1/3AC

garik1379 [7]

Answer:

AD = 1/4AC

Step-by-step explanation:

Given that,

D is the the midpoint of AB

Therefore

AB = AD + DB

Since, D is the midpoint of the AB, therefore AD = DB

AB = AD + AD

AB = 2AD (i)

Similarly, B is the midpoint of AC

Hence,

AC = AB + BC

AC = AB + AB (because B is the midpoint)

AC = 2AB (ii)

By putting the value of AB from equation i to equation ii, we get

AC = 2(2AD)

AC = 4AD

AC/4 = AD

AD = 1/4AC

3 0

3 years ago

Please help I just need the answers

bearhunter [10]

The answer to your question is 390.

6 0

3 years ago

A rucksack can hold a maximum of 6 kg.

ratelena [41]

Answer:

9 textbooks!

Step-by-step explanation:

Rucksack can hold = 6kg

Mass of textbook = 680g

So, here we need to show the division for finding the number of textbooks it can hold.

680 g in KG's = 0.68

Now, 6/0.68

That is 8.82 ( Round off = 9 textbooks)

5 0

3 years ago

Read 2 more answers