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Rufina [12.5K]
3 years ago
14

What adds to 1 and multiples to -90

Mathematics
1 answer:
wariber [46]3 years ago
6 0

Answer:

+10 and - 9

Step-by-step explanation:

If we consider the multiples of - 90

± 1, ± 2, ± 3, ± 5, ± 6, ± 9, ± 10, ± 15, ± 30, ± 90

The combination of multiples which multiply to - 90 and add to + 1 are

+ 10 and - 9

since 10 × - 9 = - 90 and 10 - 9 = 1

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If mLABC = (2x + 4)º, m_DEF= (3x – 5)°, and ABC and
Harrizon [31]

Answer:

-5x-1

Step-by-step explanation:

4 0
3 years ago
20 Point!
ludmilkaskok [199]

determinant: \sqrt{b^2-4ac}

(a) x^2+4x+5=0\\D=4^2-4\cdot1\cdot5={-4}\\D

D<0 means there are no real roots. there are two complex roots with imaginary components.

(b) D=16+20=36>0

D>0 means there are two real roots

(c) D = 20^2-4*4*25 = 0

D=0 means there is one real root with multiplicity 2


7 0
3 years ago
1. The price of a TV is $295. The price of a
TiliK225 [7]

Answer:

The price of the projector is $1,180.

Step-by-step explanation:

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5 0
3 years ago
Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water in
nadezda [96]

Answer:

11/4 seconds

Step-by-step explanation:

We can see from the given equation that the height at the diving board is h = 3.

This is because the h(t) equation has that +3 at the end, which denotes the initial height of the diver, the height when he is standing on the board before jumping.

To find where h(t) = 3 is true, we need to set h(t) equal to 3 and solve for t.

h(t) = 3

h(t) = -4t^2 + 11t + 3 = 3

-4t^2 + 11t + 3 - 3 = 0

-4t^2 + 11t = 0

t*(-4t + 11) = 0

so h(t) = 3 when t = 0 and when t = 11/4 sec

we already know that at t = 0 the height is 3, it is the initial height given from the equation, so we want to use the other solution for t.

the diver is back at the height of the diving board at t = 11/4 sec

6 0
2 years ago
Read 2 more answers
Same this for me but 6,912 and 288 please help
Svetllana [295]

Answer:

what is the question

Step-by-step explanation:

Questions for answers

4 0
3 years ago
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