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2 years ago

Please help with Geometry homework!!! Will give brainliest and 50 points to whoever answers first!!!

Mathematics

1 answer:

melomori [17]2 years ago

6 0

Answer:

x = 42.5

Step-by-step explanation:

It is said:

If two angles are not corresponding, then they make up 180°. They are more known as supplementary angles.

So the equation is:

a + b = 180°

Substitute

x + 8 + 3 x + 2 = 180°

x + 3x + 8 + 2 = 180°

4x + 10° = 180°

4x = 180° - 10°

4x = 170°

x = 170° ÷ 4

x = 42.5

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5(x+7)+2(y-1)when x=7 and y=2

balu736 [363]

Answer: 72

Step-by-step explanation:

First plug in 7 for x

5(14)

Then multiply

Next plug 2 in for y

2(1)

Multiply

Lastly, add 70 and 2 together.

72.

Hope this helps.

5 0

3 years ago

Read 2 more answers

Please answer correctly. I need this right now <3

Grace [21]

Answer:

y=9

Step-by-step explanation:

27÷3=9 per hour

Answer:

Step-by-step explanation:

27÷3=9 per hour

Answer:

49$ to rent for 5 hours

Step-by-step explanation:

9x5=49

7 0

3 years ago

Read 2 more answers

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago

The area of a square garden is 72 m2. How long is the diagonal?

Ksju [112]

The answer is 72m2 because its expression entered or try another topic

4 0

2 years ago

How do I solve part b?

Minchanka [31]

Answer:

$x \approx 1.4$

Step-by-step explanation:

This problem gives one the following equation to model the graph of a line: ( $y=x^2-2$ ). The problem asks one to find the value of (x) when (y=0). Rather than using the graph, an easier way to solve this problem is to substitute (0) in for the value of (y) and then solve for (x) using inverse operations.

$y=x^2-2\\$

Substitute,

$0=x^2-2$

Inverse operations,

$0=x^2-2\\\\2=x^2\\\\\sqrt{2}=x$

Round:

$x \approx 1.4$

As one can see on the graph, this result is very close to the value at which the graph intersects the (x) axis. Thus, one can conclude that (x) does indeed approximately equal (1.4)

6 0

2 years ago