Plz help me with number 1

Mathematics

1 answer:

BabaBlast [244]3 years ago

3 0

For figure one angle 1 is 98 and for angle 2 it is 82 THEN for figure two both angles are 64 THE LAST ONE figure 3 angle 1 is 157 and angle 2 is 23

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Find the solution to: -2x+7<-41

Karolina [17]

Answer:

x>24

Step-by-step explanation:

Solve like a normal equation :)

-2x<-48

Now we divide.

-x<-24

Note: Whenever dividing by a negative, you have to flip the sign of the inequality!

x>24

7 0

2 years ago

Read 2 more answers

Please help!!! Answer ASAP and show work!!!

n200080 [17]

Answer:

B and C. can you select more than one?

4 0

2 years ago

I need help can you help me?

UNO [17]

The remaining 2 debts have to add up to a cetain number which I represent as x.
476.25 - 93.22 - x = 0
x = 383.03
The correct two debts should add up to -383.03.
Therefore the answer is B. -66.72 and -316.31

7 0

3 years ago

Hi :) how to do question 7 ?

Gre4nikov [31]

Answer:

Step-by-step explanation:

We khow that the equation of a circle is written this way :

(x-a)²+(y-b)²=r² where (x,y) are the coordinates of the cercle's points and (a,b) the coordinates of the cercle's center and r the radius .

Our task is to khow the values of a and b :

We khow that the center is lying on the line 3x+2y=16⇒ 2y=-3x+16⇒ y= $\frac{-3}{2}$ x+8
We khow that the points P and Q are two points in the cercle
Let Ω be the center of this cercle
we can notice that : PΩ AND QΩ are both equal to the radius ⇒ PΩ=QΩ= r
So let's write the expression of this distance using vectors KHOWING THAT Ω(a,b)
Vector PΩ(a-4,b-6) and Vector QΩ(a-8,b-2)
PΩ= $\sqrt{(a-4)^{2}+(b-6)^{2} }$ and QΩ= $\sqrt{(a-8)^{2}+(b-2)^{2} }$
Let's substitute a by x and b by y
PΩ=QΩ we substitute each distance by its expression
After simplyfying the expressions we get finally : -12+8x-8y=0
now we have -12x +8x-8y=0 and the line equation 3x+2y-16=0
these are simultanious equations so after solving them we get x=3.8 wich is approximatively 4 and y=2
we substitute a by 4 and y by 2 in PΩ to get the radius
we get r = $\sqrt{(4-4)^{2}+(2-6)^{2} }$ = 4
so r²= 16
then the equation is : (x-4)²+(y-2)²=16