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ValentinkaMS [17]
3 years ago
12

Factoring a quadratic with leading coefficient greater than 1 2y^2+17y-9

Mathematics
1 answer:
abruzzese [7]3 years ago
4 0
Multiply first and last coefficients:
2*(-9) = -18

Find 2 factors which multiply to -18 and add up to 17:
18 and -1

Rewrite quadratic with 17y split up as' 18y - y', Then factor by grouping:

2y^2 - y + 18y -9 \\  \\ y(2y-1) + 9(2y-1) \\  \\ (2y-1)(y+9)
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Write two different congruent statements that indicate the triangles in each pair are congruent( Odd number questions only)
ryzh [129]

Answer:

Step-by-step explanation:

17) HI ≅ UH   ;   GH  ≅  TU     ; GI  ≅ TH

ΔHGI  ≅ ΔUTH                by Side Side Side congruent

∠G  ≅  ∠T   ; GI  ≅  TH  ; ∠GIH  ≅ ∠THU

ΔHGI  ≅ ΔUTH                by Angel Side Angle congruent

19) IJ ≅ KD   ;  IK ≅ KC   ; KJ ≅ CD

ΔIJK ≅ ΔKDC             by Side Side Side congruent

∠J ≅ ∠D    ; IJ ≅ KD ;  ∠I ≅ ∠DKC

ΔIJK ≅ ΔKDC             by Angle Side Angle congruent

7 0
3 years ago
What are the solutions of the quadratic equation below? -7x2 - 23x + 10 = 0 A. B. C. D.
-BARSIC- [3]

Answer:

There are no like terms.

7 0
2 years ago
Read 2 more answers
What is the common denominator of 3 5 and 4 10
Ahat [919]

Answer:

3 and 5, 15; 4 and 10, 20

Step-by-step explanation:

The lowest number divided by 3 and 5 is 15. Same for 4 and 10. The lowest number they can both be divided by is 20.

5 0
3 years ago
to advertise o radio,the radio station charges 32naira for the first 10 words and 2naira per word for every additional word.how
o-na [289]

Answer:

Cost for 68 words = 148 naira

Step-by-step explanation:

Given that:

Cost of first ten words = 32 naira

Cost of additional words = 2 naira per word

Let,

x be the number of additional words.

y be the total cost

y = 2x + 32

We have to find the cost for 68 words.

The cost of first 10 words will be 32 naira.

Additional words = 68 - 10 = 58 words

Putting x = 58 in the equation

y = 2(58) + 32

y = 116 + 32

y = 148

Hence,

Cost for 68 words = 148 naira

6 0
2 years ago
What horsepower is required to lift an 8,000 pound aircraft six feet in two minutes?
Dafna11 [192]

Answer:

The horsepower required is 235440 watt.

Step-by-step explanation:

To find : What horsepower is required to lift an 8,000 pound aircraft six feet in two minutes?

Solution :

The horsepower formula is given by,

W=\frac{mgh}{t}

Where, W is the horsepower

m is the mass m=8000 pound

g is the gravitational constant g=9.81

t is the time t= 2 minutes

h is the height h=6 feet

Substitute all values in the formula,

W=\frac{8000\times 9.81\times 6}{2}

W=\frac{470880}{2}

W=235440

Therefore, The horsepower required is 235440 watt.

6 0
3 years ago
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