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marissa [1.9K]
3 years ago
7

Please someone help me

Mathematics
1 answer:
nevsk [136]3 years ago
3 0

Answer:

C) <BAC ≈ <QPR and BC ≈ QR

Step-by-step explanation:

C) <BAC ≈ <QPR and BC ≈ QR

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What is the solution to the system of equations? One-fourth x minus one-half y = 8 One-half x + three-fourths y = negative 5 (–8
kiruha [24]

Answer:

Step-by-step explanation:

x/4-y/2=8, x-2y=32, x=32+2y

x/2+3y/4=-5, 2x+3y=-20, x=(-20-3y)/2

32+2y=(-20-3y)/2

64+4y=-20-3y

7y=-84

y=-12, since x=32+2y

x=32+2(-12), x=32-24=8

So the solution is the point (8, -12)

3 0
3 years ago
Read 2 more answers
Describe the location of the point having the following coordinates. positive abscissa, positive ordinate Quadrant I Quadrant II
asambeis [7]
The abscissa is the x coordinate whereas the ordinate is the y coordinate.  Both x and y are positive in the first quadrant only.
8 0
3 years ago
4. Identify the graph of the solution set of 9 &gt; 3 + 2x.
Ray Of Light [21]

Answer:

A

Step-by-step explanation:

9 > 3 + 2x

6 > 2x

3 > x

or

x < 3

and that means 3 is NOT included (otherwise there must be a smaller or equal sign). hence A is the right answer.

7 0
3 years ago
What is the equation of the line that passes through the point (5,3) and has a slope of -1/5 ?
deff fn [24]

Answer: y=-1/5x+4

Step-by-step explanation: use the equation y=mx+b (x,y)->(5,3) 5 is an x value and 3 is a y value so you want to plug in these values into the equation. 3=-1/5(5)+b. We already know our slope so just plug it into m. Now multiply -1/5 and 5 which is -1. 3=-1+b you want to isolate b so you can find the intercept. Add one to both sides so it cancels out. 4=b we found our y-intercept so we can write the equation now. y=-1/5x+4

6 0
2 years ago
You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?
s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

  = 0.693 ÷ Half Life

  = 0.693 ÷ 18

  = 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

=  (42,787) (0.0625)

= 2,674.14 g

6 0
3 years ago
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