Which is a solution to (x-3)(x 9)=-27

Answers

Answer 1

Answer: distribute
(x-3)(x+9)=-27
x^2+6x-27=-27
add 27 to both sides
x^2+6x=0
factor out x
x(x+6)=0
set each to zero
x=0
x+6=0
x=-6

x=0 and or -6

x=-6 or 0

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What two numbers multiply to 16 but add to 17

Answers

The two numbers are 16 and 1.

16+1 =17
16*1 =16

Hope this helps~

The two numbers are 1 and 16... i was confused as well but then i figured it out... it is a very good but tricky question

(Geometry), homework please help

Answers

4.That first one sounds familiar; I think I did it yesterday.

The sides don't matter (except that they're equal) so the base angle b satisfies

b + b + 38 = 180

2b = 142

b = 71

Choice a

42 + 42 + v = 180

v = 180 - 84 = 96

choice d

6. That's three acute angles, an acute triangle

7. We need to satisfy the triangle inequality which means the sum of the two smaller sides needs to be bigger than the largest

10+15>24 so choice b

SRT = RTS = 180 - STU

3x - 50 = 180 - 7x

10 x = 230

x = 23

choice a

Polygon ABCDEFGH will be dilated by a scale factor of 3.4 with the origin as the center of dilation to produce polygon A′B′C′D′E′F′G′H′. What will the length of A'H'¯¯¯¯¯¯¯ be?

Answers

A'H'¯¯¯¯¯¯¯=2*3.4 = 6.8

Answer:

6.8 units

Step-by-step explanation:

Find the area of a triangle with base of 10 inches and altitude to the base of 16 inches. 80 in² 135 in² 160 in²

Answers

Answer:

Option 1st is correct

area of a triangle is, 80 in²

Step-by-step explanation:

Area of triangle(A) is given by:

$A = (1)/(2) b \cdot h$

where,

b is the base

h is the height or altitude of the triangle respectively.

As per the statement:

Base of a triangle(b) = 10 inches and

Height = altitude(h) = 16 inches

Substitute these in [1] we get:

$A = (1)/(2)(10) \cdot 16$

⇒ $A = 5 \cdot 16 = 80$ in²

Therefore, the area of a triangle is, 80 in²

10 × 16 = 160
160 / 2 = 80

Your answer would be 80in²

Hope that helps!

(sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^

Answers

$(sin\theta+sec\theta)^2+(cos\theta+cosec\theta)^2=(1+sec\theta\ cosec\theta)^2\n\nL=sin^2\theta+2sin\theta\cdot(1)/(cos\thewta)+(1)/(cos^2\theta)+cos^2\theta+2cos\theta\cdot(1)/(sin\theta)+(1)/(sin^2\theta)\n\n=(sin^2\theta+cos^2\theta)+(2sin\theta)/(cos\theta)+(2cos\theta)/(sin\theta)+(1)/(cos^2\theta)+(1)/(sin^2\theta)$

$=1+(2sin^2\theta+2cos^2\theta)/(sin\theta\ cos\theta)+(sin^2\theta+cos^2\theta)/(sin^2\theta\ cos^2\theta)\n\n=1+(2(sin^2\theta+cos^2\theta))/(sin\theta\ cos\theta)+(1)/(sin^2\theta\ cos^2\theta)\n\n=1+(2)/(sin\theta\ cos\theta)+(1)/(sin^2\theta\ cos^2\theta)\n\n=1^2+2sec\theta\ cosec\theta+sec^2\theta\ cosec^2\theta\n\n=(1+sec\theta\ cosec\theta)^2=R$

Please help, I would greatly appreciate it :)

Answers

All answers are correct

all of the above would be correct!

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