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professor190 [17]

3 years ago

9

What type of angles are ∠2 and ∠5? Question 1 options: adjacent angles alternate exterior angles alternate interior angles corre

sponding angles none of these

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

YOU HAVE IMAGE???

Step-by-step explanation:

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I have to find the area of the figure

Digiron [165]

Answer:

. 5×17×10.3

Step-by-step explanation:

Area of ️ = 1/2bh

7 0

3 years ago

Find the limit, if it exists, or type dne if it does not exist.

Phantasy [73]

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}$

Suppose we choose a path along the $x$ -axis, so that $y=0$ :

$\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1$

On the other hand, let's consider an arbitrary line through the origin, $y=kx$ :

$\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}$

The value of the limit then depends on $k$ , which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.

8 0

3 years ago

What is the area of the figure shown?

ExtremeBDS [4]

Step-by-step explanation:

$3.5 \times 6 + (5 - 3.5) \times 6 \div 2 = \\ = 21 + 9 \div 2 = \\ = 25.5$

6 0

3 years ago

Read 2 more answers

Solve these two equations using the Addition Method (also known as the elimination method).

Brums [2.3K]

Answer:

(-2, 3)

Step-by-step explanation:

5/2(3/4x + 1/3y = -1/2)

1/2x - 5/6y = -7/2

15/8x + 5/6y = -5/4

19/8x = -19/4

x = -2

1/2(-2) - 5/6y = -7/2

-1 -5/6y = -7/2

-5/6y = -5/2

y = 3

4 0

3 years ago

What are the real zeroes of x3 + 6 x2 – 9x - 54?

o-na [289]

Answer:

Option B 3,-3,-6 is correct.

Step-by-step explanation:

We need to find real zeroes of $x^3+6x^2-9x-54$

Solving

$x^3+6x^2-9x-54\\=(x^3+6x^2)+(-9x-54)$

Taking x^2 common from first 2 terms and -9 from last two terms we get

$=(x^3+6x^2)+(-9x-54)\\=x^2(x+6)-9(x+6)\\$

Taking (x+6) common

$(x+6)(x^2-9)\\$

x^2-9 can be solved using formula a^2-b^2 = (a+b)(a-b)

$=(x+6)((x)^2-(3)^2)\\=(x+6)(x+3)(x-3)$

Putting it equal to zero,

$(x+6)(x+3)(x-3) =0\\x+6 =0, x+3=0\,\, and\,\, x-3=0\\x=-6, x=-3\,\, and\,\, x=3$

So, Option B 3,-3,-6 is correct.

7 0

3 years ago