All categories

3 years ago

I’m not sure how to do this.. can anyone help?

Mathematics

1 answer:

Pie3 years ago

5 0

Answer:

m<1 =95

m<2 = 130

Step-by-step explanation:

I take it you want the size of <1 and <2?

<ABC + 95 = 180 Same straight line.

<ABC = 180 - 95 Subtract 95 from both sides.

<ABC = 85 Combine

===========================

<1 + <ABC = 180 Interior angles on the same side of the transversal = 180

<1 + 85 = 180

<1 = 180 - 85

<1 = 95

=============================

Givens

<ABC = 85

m< 1 = 95

<ADC = 50 Given

m<2 = ?

m<2 + <ABC + m<1 + 50 = 360 The four angles of a quadrilateral add to 360

m<2 + 85 + 95 + 50 = 360

m<2 + 230 = 360

m<2 = 360 - 230

m<2 = 130

You might be interested in

Need help don’t understand

adoni [48]

there is an equation for straight lines which goes

y-y,=m(x-x,)

where( x, , y, ) is any coordinate on the line and m is the gradient

so your answer is

y-7= -2(x-4)

does this make sense?

7 0

2 years ago

Read 2 more answers

apply the function transformation of g(x)=f(x-2)+4 to f(x)=x^3-8. describe the transformation and find the final form of the fun

Fittoniya [83]

Answer:

g(x) = (x - 2)^3 - 4

Step-by-step explanation:

first, the function f is shifted to the right by 2 units and then shifted up by 4 units

4 0

3 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

Find the product in its simplest form<br><br> 2/5 * 3/7

Mamont248 [21]

Answer:

6/35

Step-by-step explanation:

I just used my calculator...

6 0

2 years ago

Read 2 more answers

What's the slope-intercept form that passes through the points (0, -1) and (1, 5)?

Rzqust [24]

Answer:

Step-by-step explanation:

$\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}$

(0, -1) ⇒ x₁ = 0, y₁ = -1

(1, 5) ⇒ x₂ = 1, y₂ = 5

So the slope:

$\bold{m=\dfrac{5+1}{1-0}=\dfrac{6}{1}=6}$

The slope-intercept form of the equation of line is y = mx + b, where m is the slope and b is the y-intercept of the line.

(0, -1) ⇒ x₀ = 0, y₀ = -1 ⇒ b = -1

Therefore:

y = 6x - 1 ← the slope-intercept form of the equation

5 0

3 years ago