All categories

3 years ago

See picture below!! :D

Mathematics

1 answer:

Gnesinka [82]3 years ago

7 0

The angles are congruent and corresponding. You can find the missing angles by subtracting the existing ones from 180.

You might be interested in

Can u solve this problem in show your work

aksik [14]

Hey there!

We can solve this problem by doing PEMDAS, which lists:

$Parentheses \\ Exponents \\ Multiplication \\ Division \\ Addition \\ Subtraction$

We have in this equation

$Parentheses \\ Exponents \\ Dvision$

Let's sp;ve this equation now!

Parentheses ↓

$(6 -3) = 3$

Exponents ↓

$3^2 = 9$

Division ↓

$\frac{36}{9} = 9$

Okay now , I show you the steps so it can be easier to solve!

$\frac{36}{(6-3)^2} \\ \\ 6-3 = 3 \\ \\ \frac{36}{3^2} \\ \\ 3^2 = 9 \\ \\ \frac{36}{9} \\ \\ \\ \\ Answer: 4$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

5 0

3 years ago

Read 2 more answers

Write a system of inequalities using the given graph

ziro4ka [17]

For dark line

(0,1)
(-1,-4)

Slope

m=-4-1/-1=5

Equation in point slope form

y-1=5x
y=5x+1

Put (0,0)

As line is dark sign is ≤

The inequality

y≤5x+1

For dashed line

(2,0)
(0,-2)

Equation in intercept form

x/2-y/2=1
x-y=2
y=x-2

Put (0,0)

0>-2

Symbol is >

Inequality is

y>x-2

7 0

2 years ago

What is the the answer

timofeeve [1]

Answer: x=r^2 + 10 r + 25

7 0

2 years ago

(6+3)-+8-7)= (1+4)+(10-8)= (3-3+(9+1)= (10-2)+(10-8)= 2+(0+1+2+3)= (10-1-2-3)+4=

JulijaS [17]

Answer:

6+3-8-7 = 8

1+4+10-8 = 7

3-3+9+1 = 10

10-2+10-8 = 10

2+0+1+2+3 = 8

10-1-2-3+4 = 8

Step-by-step explanation:

i don't know if these are supposed to be easy or not lol but thanks for the points :p

6 0

2 years ago

Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat

Rashid [163]

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

$=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t)$

$= - \sec^2 t \frac{dt}{dx}$

$= - \sec^2 t \left(\frac{1}{\cos t}\right)$

$= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \ \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$

8 0

3 years ago