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kati45 [8]
2 years ago
6

Identify the true statements if ∆XWT - ∆YAQ.

Mathematics
1 answer:
Natali [406]2 years ago
5 0

Answer:

Step-by-step explanation:

Its A and c

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Please help and thank you!
Rainbow [258]

Answer:

Its d i took the class

Step-by-step explanation:

7 0
3 years ago
PLEASE HELP!!! WILL GIVE BRAINLIEST!!!
777dan777 [17]

Answer: (2,2) and (-5,16)

Step-by-step explanation:

Here we have both Line (Linear Function) and Parabola (Quadratic Function)

So I am gonna write these equations here,

y=x^2+x-4\\y=-2x+6

The first equation has Parabola graph (Since it's second degree.)

and the second equation has line graph.

To find the intersection, you have to substitute either -2x+6 in first equation (Quadratic) or x^2+x-4 in second equation (Linear)

For me, I am going to substitute x^2+x-4 in y=-2x+6.

x^2+x-4=-2x+6

Now solve the equation and find the value of x.

Since it's Quadratic Equation (Because there's x^2) I'd move -2x+6 to the left side.

x^2+x-4+2x-6=0 Finish things here (Subtract and Addition)

x^2+3x-10=0 What two numbers multiply to 10? Find the factors of 10, that are [1 and 10] and [2 and 5]

Now think about it, do you think that if 1 and 10 subtract or even addition, do you think that it'd be 3? No, of course not.

So 2 and 5 is right.

(x-2)(x+5)=0 (5-2 = 3) and (5*(-2) = -10)

Then we get both x, x=2,-5

However, this is not it. You have to substitute both x in Linear Equation.

Substitute x = 2 in y=-2x+6

y=-2(2)+6\\y=-4+6\\y=2

Order = (2,2)

Then substitute x = -5 in y=-2x+6

y=-2(-5)+6\\y=10+6\\y=16

Order = (-5,16)

So the intersections are both (2,2) and (-5,16) as shown in graph below.

4 0
3 years ago
Write the first five digits of 1/7 in base 9 expression
kaheart [24]

Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:

1/9 = 7/63

2/9 = 14/63

while

1/7 = 9/63

Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,

1/7 = 1/9¹ + 2/63.

We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and

1/9² = 7/567

2/9² = 14/567

3/9² = 21/567

while

2/63 = 18/567

Then

2/63 = 2/9² + 4/567

so

1/7 = 1/9¹ + 2/9² + 4/567

Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and

1/9³ = 7/5103

2/9³ = 14/5103

3/9³ = 21/5103

4/9³ = 28/5103

5/9³ = 35/5103

6/9³ = 42/5103

while

4/567 = 36/5103

so that

4/567 = 5/9³ + 1/5103

and so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/5103

Next, LCM(5103, 9⁴) = 45927, and

1/9⁴ = 7/45927

2/9⁴ = 14/45927

while

1/5103 = 9/45927

Then

1/5103 = 1/9⁴ + 2/45927

so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927

One last time: LCM(45927, 9⁵) = 413343, and

1/9⁵ = 7/413343

2/9⁵ = 14/413343

3/9⁵ = 21/413343

while

2/45927 = 18/413343

Then

2/45927 = 2/9⁵ + remainder

so

1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/9⁵ + remainder

Then the base 9 expansion of 1/7 is

0.12512..._9

8 0
2 years ago
Helpppp meeeee pleaseee
Sedbober [7]
18 I’m pretty sure that is the answer that
U want
4 0
3 years ago
Find the area of an equilateral triangle (regular 3-gon) with 6-inch sides. Round your answer to the nearest hundredth.
Rainbow [258]
Area=9ROOT3 inches
Anyway wanna talk?!

7 0
2 years ago
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