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3 years ago

9

QUICK !!! FAST 15 POINTS I NEED HELP!!!!!

2 answers:

Firdavs [7]3 years ago

8 0

Answer:

ECB

Step-by-step explanation:

I noticed they looked exactly the same.

Lana71 [14]3 years ago

3 0

Answer:

It should be ΔABD, ΔECD, and ΔECD

Step-by-step explanation:

Since ΔABE is a right triangle, one angle must be 90° with a square to signify it.

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Traveling at a average speed of 50 miles per hour the trip from point a to be is 2 hours traveling at a speed of 20 miles per ho

dem82 [27]

What's your question..?

7 0

3 years ago

Somebody please help me: Evaluate the expression for x = -1/4

Dahasolnce [82]

Answer:

There are no expressions provided in the question.

4 0

3 years ago

Using the binomial theorem , obtain the expansion of :

andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1 6 15 20 15 6 1

Decreasing powers of 1 from $1^{6}$ to $1^{0}$

Increasing powers of 3x from $(3x)^{0}$ to $(3x)^{6}$

$1+3x)^{6}$

= 1. $1^{6}$ $(3x)^{0}$ + 6. $1^{5}$ $(3x)^{1}$ + 15. $1^{4}$ $(3x)^{2}$ + 20. $1^{3}$ $(3x)^{3}$ + 15.1² $(3x)^{4}$ + 6. $1^{1}$ $(3x)^{5}$ + 1. $1^{0}$ $(3x)^{6}$

= 1 + 18x + 135x² + 540x³ + 1215 $x^{4}$ + 1458 $x^{5}$ + 729 $x^{6}$

--------------------------------------------------------------------------------------

$(1-3x)^{6}$

= 1. $1^{6}$ $(-3x)^{0}$ + 6. $1^{5}$ $(-3x)^{1}$ + 15. $1^{4}$ $(-3x)^{2}$ + 20. $1^{3}$ $(-3x)^{3}$ + 15.1² $(-3x)^{4}$ + 6. $1^{1}$ $(-3x)^{5}$ + 1. $1^{0}$ $(-3x)^{6}$

= 1 - 18x + 135x² - 540x³ + 1215 $x^{4}$ - 1458 $x^{5}$ + 729 $x^{6}$

----------------------------------------------------------------------------------

Collecting like terms from both expressions

$(1+3x)^{6}$ + $(1-3x)^{6}$

= 2 + 270x² + 2430 $x^{4}$ + 1458 $x^{6}$

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1 5 10 10 5 1

Decreasing powers of 1 from $1^{5}$ to $1^{0}$

Increasing powers of 2x from $(2x)^{0}$ to $(2x)^{5}$

$(1+2x)^{5}$

= 1. $1^{5}$ $(2x)^{0}$ + 5. $1^{4}$ $(2x)^{1}$ + 10. $1^{3}$ $(2x)^{2}$ + 10. $1^{2}$ $(2x)^{3}$ + 5. $1^{1}$ $(2x)^{4}$ + 1. $1^{0}$ $(2x)^{5}$

= 1 + 10x + 40x² + 80x³ + 80 $x^{4}$ + 32 $x^{5}$

8 0

3 years ago

Identify the equivalent expression for (m^1/3 m^1/5)^0

Lostsunrise [7]

Answer:

1.

Step-by-step explanation:

The thing is, x^0 is always 1. So even if you simplify, it will still be 1.

$(m^\frac{1}{3}*m^\frac{1}{5})^0\\(m^\frac{8}{15})^0\\\text{But when put to the power of 0,}\\1.$

4 0

3 years ago

Read 2 more answers

Find the equation,in standard form, of the line passing through the point (2,-3) and (4,2)

VLD [36.1K]

The standard form:

$Ax+By=C$

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - point

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (2, -3) and (4, 2). Substitute:

$m=\dfrac{2-(-3)}{4-2}=\dfrac{5}{2}$

$y-(-3)=\dfrac{5}{2}(x-2)$

$y+3=\dfrac{5}{2}(x-2)$ <em>use distributive property</em>

$y+3=\dfrac{5}{2}x-5$ <em>multiply both sides by 2</em>

$2y+6=5x-10$ <em>subtract 6 from both sides</em>

$2y=5x-16$ <em>subtract 5x from both sides</em>

$-5x+2y=-16$ <em>change the signs</em>

$\boxed{5x-2y=16}$

8 0

3 years ago

Read 2 more answers