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tatyana61 [14]
3 years ago
10

I need help with this​

Mathematics
1 answer:
Ulleksa [173]3 years ago
8 0
No, tahts the answer
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50 points! Due in 20 minutes! So hurry ! no random answers please. Will mark brainliest:)
olga2289 [7]
The new coordinates are (3, 1)
5 0
3 years ago
Please help me solve this for 15 points! <br><br>-x2 + 4x - 54 = 0​
Mazyrski [523]

Answer:

x1 =2-5i*sqrt(2)

x2 =2+5i*sqrt(2)

Step-by-step explanation:

-x^2 +4x-54=0 (quadratic equation)

a=-1, b=4, c=-54

x1=(-b+sqrt(b^2-4ac))/2a

x1=(-4+sqrt(4^2 - 4*(-1)(-54))/2*(-1)

x1=(-4+sqrt(16-216))/(-2)

x1 =(-4+sqrt(-200))/(-2)

x1 =(-4+sqrt(200i^2))/(-2) i^2=-1

x1 =(-4+sqrt(100*2*i^2))/(-2)

x1 =(-4+10i*sqrt(2))/(-2)

x1 =2-5i*sqrt(2)

x2 =(-b-sqrt(b^2-4ac))/2a

x2 =(-4-10i*sqrt(2))/(-2)

x2 =2+5i*sqrt(2)

7 0
3 years ago
Read 2 more answers
H Quarshie
Yuki888 [10]

Answer:

28

Step-by-step explanation:

f(6) = -3 * 6 + 10 = -18 + 10 = -8

g(f(6)) = g(-8)

g(-8) = (-8)^{2} + 3 * (-8) - 12 = 64 - 24 - 12 = 64 - 36 = 28

6 0
3 years ago
What are the possible lengths for x, the third side of a triangle, if two sides are 21 and 10
SashulF [63]
To solve we need to use Pythagoras theorem ( a^{2} +  b^{2} =  c^{2} )

There are 2 possible lengths for x: hypotenuse or one of the 2 shorter sides.

Hypotenuse:
10^{2} + 21^2 = x^{2}
100+441= x^{2}
\sqrt{541} =  \sqrt{ x^{2} }
23.3≈x

Shorter leg:
10^2 + x^2=21^2
x^2= 21^2-10^2
x^{2}= 441-100 
\sqrt{ x^{2} } = \sqrt{341}
x≈18.47
8 0
3 years ago
Please help as soon as possible
GenaCL600 [577]

Answer:

The first is -4 and the second is 10

Step-by-step explanation:

7 0
2 years ago
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